Tangent of a curve
1.  Tangent: Means touching line
2.  Tangent: Let y  =  f (x) be a curve and P be a point on the curve.
           If Q (P) is a point on the curve, then is called a secant line. If the secant line  approaches the same limiting position as Q moves along the curve and approaches to P from either side, then limiting position is called a tangent line to the curve at P. The point P is called point of contact of the tangent line to the curve y = f(x).
Note: If P is a point on the curve y = f(x) then there exists all most one tangent at P to the curve y = f(x).
Normal of a curve
Normal: Let y = f(x) be a curve and P be a point on the curve. The line passing through P and perpendicular to the tangent y = f(x) at P is called the normal to the curve y = f(x) at P.
Gradient of a curve: Let y = f(x) be a curve and P be a point on the curve. The slope of the tangent to the curve y = f(x) at P is called gradient to the curve at P.

Angle between two curves
*  The acute angle which two curves intersect at a point is defined as the angle between their tangents at the point of intersection.
*  If the angle at a point of intersection is a right angle then the two curves are said to cut each other orthogonally at the point of intersection.
*  If the angle at a point of intersection is zero then the two curves are said to touch each other at the point of intersection.
*  If m1 is the slope of the tangent to the curve f(x) and m2 is the slope of the tangent to the

             curve g(x) such that 1 + m₁ m₂ 0 then tan  
*  If 1 + m₁m₂ = 0 m₁m₂ =  -1 then the two curves cut each other orthogonal at P.
*  If m₁ = m₂ then the two curves touch each other at P. In this case the two curves have a common tangent and a common normal at P.
*   Let us consider a curve whose equation is y =  f(x).
*   Let P (x₁, y₁) be a point on the curve.

 Let P (x1, y1) be a point on the curve.
*  Let the tangent at P be not perpendicular to X -axis.Let the tangent meet the X -axis in T.
*  Draw PN perpendicular to the tangent at P to meet  in N. Draw   
*  The length of the segment PT is called the length of the tangent at P.
*  The length of the segment PN is called the length of the normal at P.
*  TQ, The projection of PT on OX, is called the length of the sub tangent.
*  QN, The projection PN on OX, is called the length of the sub normal.

Let f: A R be a function, then
*  f is said to be monotonically increasing on A if x₁, x₂ A, x1 < x₂ f(x1) ≤ f(x2).
*  f is said to be strictly increasing on A if x1, x2 A, x1 < x2 f(x1) < f (x2)
*  f is said to be monotonically decreasing on A if x1, x2  A, x1 < x2  f(x1) ≥ f(x2)
*  f is said to be strictly decreasing on A if x1, x2  A,  x1 < x2  f (x1) > f(x2)
*  f is said to be monotonic on A if f is either monotonically increasing or decreasing.
Let f be a function defined on a neighbourhood A of a real number "a"
Then f is said to be locally 
       Increasing at 'a' if
*  x  A,  x < a  f(x) < f(a)
* x  A,  x > a  f(x) > f(a)
        Decreasing at 'a' if
*  x  A, x < a  f(x) > f(a)
*   x  A, x > a  f(x) < f(a)

 Let f be a function defined on a neighbourhood A of a real number 'a'
If f'(a) > 0 then f is increasing at 'a'.
If f' (a) < 0 then f is decreasing at 'a'.
Let f :  A  R be a function and a  A. Then f is said to be stationary at 'a' if f'  (a) = 0.
f(a) is also called stationary value of f at 'a' and the point ( a, f(a) ) is called stationary
point (or) turning point of f.
Let f : A  R be a function and l  f(A). Then 'l' is said to be
The maximum (or) greatest value of f in A  if  f (x) ≤ l  x  A
The minimum (or) least value of f in A  if  f (x) ≥ l  x  A
Note: A function need not have maximum (or) minimum values in a set.
Let f be a function defined on a neighbourhood A of a real number 'a'
Then f is said to have
  

 Note:  f(a) is called relative maximum (or) relative minimum of f at 'a'
Let f be function defined on [a, b]
       The maximum of all relative maximum values of f on [a, b] is called absolute maximum of f on [a, b]
       The minimum of all relative minimum values of f on [a, b] is called absolute minimum of f on [a, b]
Note:  The relative maximum and the relative minimum values of f are called EXTREME Values of f.

 

Velocity of the particle at the time 't' is :
Note (1): If v > 0 then s is increasing and the particle is moving from left to right.
Note (2): If v < 0 then s is decreasing and the particle is moving from right to left.
Note (3): If v = 0 then the particle comes to rest.
Acceleration :
         The rate of change of velocity i.e. is called the acceleration of the particle at the instant. It is denoted by 'a'.
  Acceleration of the particle at the time 't' is:   =  
Angular velocity : 
         Let 'O' be a fixed point in a plane and   be a fixed ray. Let P be the position of a particle on a curve C at the time 't' such that  
If t is any change in t and θ is the corresponding change in θ then the rate of change in θ is   which is called the angular velocity of the particle about O. It is denoted by

Angular Acceleration :
          The rate of change of angular velocity is   which is called the angular acceleration of the particle.
                                 

 is called the percentage error in x.

 The following formulae will be very useful in solving the problems.

Circle: If r is the radius, d is the diameter, P is the perimeter (circumference) and A is the area of a circle then        
           
Cube:  If x is the side, S is the surface area and V is the volume of a cube then     
                
Cone:  If 'r' is the base radius, 'h' is the height, 'l' is the slant height,   is the vertical angle,  is the semi-vertical angle, L is the lateral surface area, S is the total surface area and V is the volume of a (right circular) cone then
                            

Cylinder:  If r is the radius, h is the height, L is the lateral surface area, S is the total surface area and V is the volume of a (right circular) cylinder then   

Sphere:  If r is the radius, S is the surface area and V is the volume of a sphere then    

Simple pendulum:  If l is the length, T is the period of oscillation of a simple pendulum and y is the acceleration due to gravity, then    

Sector:  If r is the radius, l is the length of the arc and   is the angle,  p is the Perimeter and A is the area of a sector then        

 Today, not only Mathematics but many other subjects such as physics, chemistry and economics are enjoying the fruits of calculus. The credit goes to the great English Mathematician Sir Isaac Newton and the noted German Mathematician G.W. Libnitz, who independently invented calculus around seventeenth century. After the advent of calculus many mathematicians contributed for further development of mathematics. Cauchy gave the foundation of calculus. He used D' Alembert's limit concept to define the derivative of a function. He wrote

Geometrically, f'(x) is interpreted as the slope of the curve at the point (x, f(x)). The line through (x, f(x)) which has this slope is called tangent at (x, f(x)). If there is no tangent at a certain point then the function is not differentiable at that point.
* Right hand derivative of f(x) at x = x0 is denoted by f'(x0+) and is defined as
       and 
    Left hand derivative of f(x) at x = x0 is denoted by f'(x0-) and is defined as 

    If f'(x0+) = f'(x0-) then f(x) is differentiable at x = x0.
* If a function is differentiable on [a, b] then f(x) is differentiable at every 

* Every differentiable function is continuous but not the converse.
* If y = f(x) is a differentiable function such that z = f'(x) is also differentiable then the second derivative of y = f(x) is denoted by y2(x), f''(x)or and difined as

 If f'(x) = 0 at every point of a certain interval, then f(x) is constant on that interval.
* If y = f(x) satisfies the equation F (x, y) = 0 then to find , when y is differentiable,
differentiate F w.r.t. x considering y as a function of x and solve for .
* If x = f(t) and y = g(t) are two differentiable functions then 

*  Use of logarithms will be of great help in finding the derivative of the function of the form    
              (i)  y = f(x)g(x)

 If y = f(x) and z = g(x) then  
* If y = [f(x)]y then   =
* If f(x + y) = f(x).f(y) and f'(0) exists then f'(x) = f(x) f'(0).

I. Concepts and formulae:
1) Let f(x) be a function in x. Let it be denoted as y = f(x) where 'x' is an independent variable and 'y', a dependent variable.
         Let x be a small change in 'x'. Correspondingly, let y be the small change in 'y'. Therefore, we have

2) The other notation which is in common use is as follows:
       f'(x) =    where 'h' is a small change in 'x'.
3) The derivative of a function at x = a is defined as
       f'(a) =  
4) For x  R, f(x) =  is not differentiable at zero and is differentiable for x ≠ 0.
5) The derivative of a constant function is zero.
6) If f(x) is differentiable at x = a, then f(x) is continuous at x = a. The converse of this need not be true.
7) Some of the standard derivatives are given below:

\

 

The concept of 'Limit' is fundamental to the study of Calculus which is an important branch of advanced Mathematics. The precise definition of 'Limit' is difficult to understand without looking at the Geometrical interpretation of 'Limit' and some examples.
Geometrical Interpretation of Limit : When a regular polygon of 'n' sides is inscribed in a circle it appears that as 'n' increases infinitely there is possibility of the polygon coinciding with the circle. But theoretically for any large value of 'n' the polygon never coincides with the circle. In such a case, we say that the limiting shape of the polygon is circle as 'n' increases infinitely and by notation it is denoted as      
   (Regular polygon of 'n' sides inscribed in a circle) = circle.
  Limit of a function: Consider the function  

 Now as x approaches the value '2', values of the function are given in the form of a table.
     Here f(x) is not defined at x=2 but we observe that as x approaches to '2' the value of f(x) approaches to '4'. In this case we say that limiting value of f(x) is '4' as x approaches to '2'.
This is denoted as

Mathematical definition of Limit
    If 'f' is a function of a variable 'x' and defined in a deleted neighbourhood of 'a', the number 'l ' is said to be a limit of f(x) as x→a, if for any arbitrarily chosen positive number δ, however small it may be, there exists a corresponding number ε greater than zero such that |f(x) -1| < ε for all values of 'x', for which 0 < | x-a | < δ. It is denoted by

    In other words, a function f(x) is said to tend to a limit 'l ' as 'x' tends to 'a', if the difference between f(x) and l can be made as small as we please by taking x sufficiently nearer to a.
                            We write it as
Note: i) The function need not be defined at x = a.
ii) The value of f(x) at x = a has no effect on the existence or non existence of the limit.
Infinite Limit: Let f be a function defined in a deleted neighbourhood of 'a'. If for every K > 0 (however large), there exists a, δ > 0 such that that f(x) > K whenever 0 < |x-a| < δ and is written as  

Indeterminate Forms: While finding the limiting values of certain functions, forms such as may be encountered. 

Since for the same form in different cases, different limiting values may be obtained, these forms are called indeterminate forms.
     Methods like substitution, rationalisation, formula application may be employed to eliminate these forms.

   If all the points of a straight joining any two points on the surface lie on the surface then the surfae is called a plane.
Let the first degree equation in x,  y,  z  be  ax  +  by  +  cz  +  d  =  0   ......... ( 1 )
Let A  ( x₁,  y₁,  z₁,  )  and
       B  ( x₂,  y₂,  z₂  ) be any two points in surface represented by ( 1 )
then       ax₁  +  by₁  +  cz₁  +  d  =  0    .......( 2 )
              ax₂  +  by₂  +  cz₂  +  d  =  0    ......( 3 )
 
Let     be any point on   
 Let  C  divide  in the ratio m  :  n

 

We know that how to find the angle between two lines in a plane.
* If two lines in the space are intersecting then there exists a unique plane passing through those two lines.
* In this case the angle between the lines is similar to that of in the two dimensional geometry.
* Now we define the angle between two non-intersecting lines.
* The angle between two non-intersecting lines is defined as the angle between the two lines drawn parallel to them through any point in the space.
* If a given directed line (ray) makes angles α, β, γ with positive direction of the axes of x, y and z respectively then cos α, cos β, cos γ are called the direction cosines (d.c s) of the line and they are denoted by l, m, n.
                             i.e. l = cos α            m = cos β               n = cos γ
* If l, m, n are direction cosines of a line then -l, -m, -n are also direction cosines of the line. Usually we take one trial of direction cosines l, m, n and these are denoted by the ordered triple (l, m, n).
Since the line  makes with the positive directions of x - axis, y- axis, z-axis, the angles 0°, 90°, 90° respectively.
* Then the direction cosines of x-axis are (cos0°, cos90°, cos 90°) i.e. (1, 0, 0) Similarly (0, 1, 0) and (0, 0, 1) are the direction cosines of y and z - axes respectively.

Let α,β,γ be the angels made by the directed line  with the positive direction of x-axis, y-axis, z - axis respectively such that l = cos α, m = cos β, n = cos γ
Let (x, y, z) be the coordinates of P
               Let M be the projection of P on x - axis.
                                                 Then OM = x

                                                        x  =  l r
                                       Similarly     y  =  mr
                                       and               z  =  nr
                                                      P = (lr, mr, nr)
Let P be a point on the line Parallel to the given line and passing through the origin such that OP = 1 
                                            Then, P = (l,m,n)
                                            Now OP = 1

²+m²+n² = 1
 cos² α + cos² β +  cos² γ = 1 Where α, β, γ are the angles made by adirected line with the positive direction of coordinate axes.
Three real numbers a,b,c are said to be direction ratios (or) direction numbers of a line if a:b:c = l:m:n where (l,m,n) are d.c.'s of the line.
                                               a : b : c    =     l : m : n

 

 We know that, in analytical geometry of two dimensions the position of a point is determined w.r.t. two axes of reference.
But in the Space it is not possible to determine the point w.r.t. two axes.
Thus to locate the position of a point in Space another third axis is required besides the two axes.
Such a coordinate system in Space is called a three dimensional system.
 If xox1, yoy1 are two mutually perpendicular coordinate lines, intersecting at o then three exists a unique plane passing through these two lines.
 If zoz1 is a coordinate line perpendicular to this plane and passing through o then o is called origin.
 These three mutually perpendicular lines xox1, yoy1, zoz1 are called x - axis, y - axis, z - axis respectively.     

The plane passing through xox₁ and yoy1 is called xy - plane.
The plane passing through yoy₁ and zoz1 is called yz - plane
The plane passing through zoz₁ and xox1 is called zx - plane
              x - axis, y - axis and z - axis are called coordinate axes and the planes   xy  plane, 
yz - plane, zx - plane are called coordinate planes. Such a system of coordinate axes is called rectangular cartesian coordinate system.

Coordinate of a point in Space
Let P be any point in Space
Let A,  B,  C be the projections of P on X ,Y, Z coordinate axes.
Let x be the X - coordinate of A, y be the Y-coordinate of B and z be the Zcoordinate of C.
Then x,  y,  z are called the X,  Y,  Z coordinates of the point P and is denoted by the ordered trial, P  =  ( x,  y,  z ).

Let (  x,  y,  z )  be a given coordinate trial.
Let A be the point on X-axis with x as its X - coordinate, B be the point on Y - axis with y as its Y - coordinate and C be the point with z as its Z - coordinate on Z - axis.
Through the point A draw plane perpendicular to  . Now A will be the foot of the perpendicular drawn to   from any point in this plane. So x is the X - coordinate of any point in that plane.
Similarly, draw plane perpendicular to  and passing through B. Then y is Y - coordinate of every point in this plane.
Draw another plane passing through C and perpendicular to . Every point on this plane will have z for its Z - coordinate.
These three mutually perpendicular lines intersect in a unique point P in Space and the coordinates of P are given by the ordered trial (x, y, z).
The three coordinate planes divide the Space into eight equal parts. Each part is called an Octant.
The signs of the coordinates of a point determine the octant in which it lies.
The following table shows the signs of the coordinates of points in the eight octants.

 Two lines together (whether they are parallel, intersect each other or coincide) are called Pair of straight lines.
Combined Equation Of Pair Of Lines:
           If the equations of two straight lines separately are L₁ = 0 and L₂ = 0, then the
combined equation of that pair of lines is given by L₁ L₂ = 0
Reason: (i) All the points of L₁ = 0, satisfy the equation L₁ L₂ = 0
(ii) All the points of L₂ = 0, satisfy the equation, L₁ L₂ = 0
(iii) Those points which are not on L₁ = 0 or L₂ = 0, donot satisfy the equation L₁ L₂ = 0
eg: 1. The combined equations of the lines 2x + 3y + 5 = 0 and 3x - 4y + 7 = 0 is
           (2x + 3y + 5)  (3x - 4y + 7) = 0
2. The combined equation of the lines 4x + y = 0 and 2x + 5y = 0 is (4x + y) (2x + 5y) = 0
Result: The combined equation of two lines
L₁ = l₁x + m₁y = 0
L₂ = l₂x + m₂y = 0,

the equation ax² + 2hxy + by² = 0 represents pair of (real) lines
Important Result: If the scopes of the two (non - vertical) lines represented by ax² + 2hxy + by² = 0 are m₁, m₂ respectively, then we have

Angle between pair of lines: If the equation ax² + 2hxy + by² = 0 represents pair of lines and 'θ' is acute angle between them,

Note: 1. When θ = 90°, a + b = 0
2.  When θ  =  0°,  h² = ab
Bisectors of Angles:
Result: We can show that, the equations to the bisectors of angles between the lines
a₁x + b₁y + c₁ = 0
And a₂x + b₂y + c₂ = 0 are

eg: Find the bisectors of angles between the two given lines
3x + 4y + 1 = 0 and 5x + 12y + 3 = 0
Sol: The equations of bisectors of angle are 

 By simplifying, we can get the equations of bisectors of the angles between given pair of lines.
Important Result: The combined equation of pair of bisectors of angle between the pair of lines ax² + 2hxy + by² = 0 is given by
h(x² − y²) − (a − b)xy = 0
5555555555555555555

Important Results:
1. If the equation S ≡ ax² + 2hxy + by² + 2gx + 2fg + c = 0 represents a pair of lines, then
(i) ∆ ≡ abc + 2fgh − af2 − bg² − ch² = 0
              And
(ii) f² ≥ bc, g2 ≥ ca, h² ≥ ab
2. If S ≡ ax² + 2hxy + by² + 2gx + 2fy + c = 0
represents a pair of parallel lines, then
(i) h² = ab
(ii) af² = bg²

3. The point of intersection of fair of lines,
S ≡ ax² + 2hxy + by² + 2gx + 2fy + c = 0

Curves represented by the general equations of 2nd degree in x, y:
         Depending on some conditions, the general equation of 2nd degree in x, y
         S ≡ ax² + 2hxy + by² + 2gx + 2fy + c = 0
         may represent, a pair of lines or a circle or a conic or may not represent any real point also.
        Now, we will discuss the method of finding out the equation to the pair of lines formed by joining the origin to the points of intersection of given curve ax² + 2hxy + by² + 2gx + 2fy + c = 0 and a line lx + my = 1

Since the required pair of lines pass through origin and satisfy the common point of given curve and lines. We get the equation of required pair of lines by homogenising the curve using the given line.
eg: Find the combined equation of pair of lines, obtained by joining the origin to the point of intersection of x² + y² = 1 and x + y = 1.
Solution:

If a line makes an angle θ with x - axis in the positive direction, then θ is called inclination of the line and 'tan θ' is called slope of the line and is denoted by m.
* Slope of x - axis or any line parallel to it is "Zero".
* Slope of y - axis or any line parallel to it is "tan 90o" not defined.
* Slope of a line passing through two points
    (x₁, y₁) and (x₂, y₂) is   x₁ ≠ x₂
* Two lines are parallel if their slopes are equal.
* Two lines are perpendicular if product of their slopes is equal to '-1'.
* Equation of x - axis is y = 0
    Equation of y - axis is x = 0
    Equation of line parallel to x - axis is y = K (K = constant).
    Equation of line parallel to y - axis is x = K (K = constant).
Equation of line passing through (x₁, y₁) and having slope 'm' is y - y₁ = m(x - x₁)
* Equation of line passing through (x₁, y₁) and (x₂, y₂) is 
                         (y - y₁) (x₂- x₁) = (y₂ - y₁) (x - x₁).
* Equation of line having slope 'm' and Y - intercept 'c' is y = mx + c.

* Equation of line making X - intercept 'a' and y-intercept 'b' is 
* ax + by + c = 0 is the general form of equation of a straight line with slope  and intercepts ,   respectively on coordinate axes.
* Equation of line with inclination θ and passing through (x₁, y₁) is 
             (symmetric form) where θ  (0, Π/2) U (Π/2, Π)
* x = x₁ + r cos θ, y = y₁ + r sin θ are the parametric equations of line passing through (x₁, y₁).

 The point P(x, y) is a general point on the Cartesian plane. The x and y are denote the abscissa and ordinate of a point respectively. The co-ordinates of the general point P, both x and y are variables therefore the point P is also called a variable point. When a point P moves under certain conditions then the path traced out by the point P is called the locus of the point. The locus of a variable point P (x, y) is called a curve. In coordinate geometry, we deal with two types of fundamental problems.
1. Given locus of a variable point (geometrical condition), to find the corresponding equation (algebraic relation)
2. Given an equation, to find the corresponding curve.
Definition of locus: A locus is the set of points (and only those points) that satisfy the given consistent geometric condition(s).
      From the above definition, it follows that:
(i) Every point satisfying the given conditions is a point on the locus.
(ii) Every point on the locus satisfies the given conditions.
Note: Locus is a Latin word which means 'location' or 'place'.The plural of locus is loci. A set of geometric conditions is said to be 'consistent', if there is atleast one point satisfying the set of conditions.
      For example, when A = (2, 0) and B = (5, 0), the condition 'the sum of the distances of a point from A and B is equal to 3' is consistent. Whereas the condition 'the sum of distances of a point Q from A and B is equal to 2' is not consistent. The locus of the point equidistant from three non collinear points. The locus consists of only one point. Which is the circumcentre of the triangle formed the three non-collinear points.
Equation of Locus: The equation of a locus we mean an algebraic description of the locus. It is obtained by translating the geometric conditions satisfied by the points on the locus, into equivalent algebraic conditions.
Procedure for find locus

      An equation of a locus is an algebraic description of the locus. This can be obtained in the following way.
 I) P(x, y) or P(h, k) or P(x1, y1) or P(α, β) be a point on the locus. Select point P suitable to geometric condition(s).
 II) Write the geometric condition(s) to be satisfied by P in terms of an equation or inequation in symbols.
 III) Apply the proper formula of co-ordinate geometry and translate the geometric conditio (s) into an algebraic equation.
 IV) Simplify the equation so that it is free from radicals.

                                                                 

The point P(x, y) is a general point on the Cartesian plane. The x and y are denote the abscissa and ordinate of a point respectively. The co-ordinates of the general point P, both x and y are variables therefore the point P is also called a variable point. When a point P moves under certain conditions then the path traced out by the point P is called the locus of the point. The locus of a variable point P (x, y) is called a curve. In coordinate geometry, we deal with two types of fundamental problems.
1. Given locus of a variable point (geometrical condition), to find the corresponding equation (algebraic relation)
2. Given an equation, to find the corresponding curve.
Definition of locus: A locus is the set of points (and only those points) that satisfy the given consistent geometric condition(s).
 From the above definition, it follows that:
(i) Every point satisfying the given conditions is a point on the locus.
(ii) Every point on the locus satisfies the given conditions.
Note: Locus is a Latin word which means 'location' or 'place'.The plural of locus is loci. A set of geometric conditions is said to be 'consistent', if there is atleast one point satisfying the set of conditions.
      For example, when A = (2, 0) and B = (5, 0), the condition 'the sum of the distances of a point from A and B is equal to 3' is consistent. Whereas the condition 'the sum of distances of a point Q from A and B is equal to 2' is not consistent. The locus of the point equidistant from three non collinear points. The locus consists of only one point. Which is the circumcentre of the triangle formed the three non-collinear points.
Equation of Locus: The equation of a locus we mean an algebraic description of the locus. It is obtained by translating the geometric conditions satisfied by the points on the locus, into equivalent algebraic conditions.
Procedure for find locus

An equation of a locus is an algebraic description of the locus. This can be obtained in the following way.

 I) P(x, y) or P(h, k) or P(x1, y1) or P(α, β) be a point on the locus. Select point P suitable to geometric condition(s).
II) Write the geometric condition(s) to be satisfied by P in terms of an equation or inequation in symbols.
III) Apply the proper formula of co-ordinate geometry and translate the geometric conditio (s) into an algebraic equation.
 IV) Simplify the equation so that it is free from radicals.

                                                                 

1. Find the equations of the tangent and normal to the curve y = x³ - 3x² - x + 5 at (1, 2)
sol:    Given Curve :  y  =  x³ - 3x² - x + 5
               Differentiating w.r.t.  'x'
          Slope     =  3x² - 6x - 1
            Given point : P(1, 2)   
       Slope at P (1, 2) :   P =  3(1)² - 6(1) -1  =  - 4
          
                                ⇒   y - 2 = - 4 (x - 1)
                                 ⇒  y - 2 = - 4 x + 4
                                  ⇒ 4x + y - 6 = 0

                              ⇒     y - 2 =    (x - 1)
                              ⇒  4y - 8  =  x - 1
                              ⇒    x - 4y + 7 = 0

2.  Show that the curves 6 x² - 5x + 2y = 0 and 4 x² + 8y² = 3 touch each other at 
sol :  Given curves : 6x² - 5x + 2y = 0  ----- (1)
                                   4x² + 8y² = 3 --- (2)
                    Given point :  P 
                From (1) ,  slope : 12x - 5 + 2   =  0
                                              
           

                                                ∴     m₁ = m₂
                          Given two curves touch each other at  .

3.  Find the angle between the curves xy = 2 and x² + 4y = 0.
Ans:               Given lines xy = 2  -------->  (1)
                              x² + 4y = 0   ---------> (2)
                        From (1) and (2)
                      ⇒      x² + 4 () = 0
                     ⇒       x³ + 8 = 0
                     ⇒    x³  =  -8
                           ∴    x = -2
        From (1) :  y =        
                       ⇒   y  =  -1
      ∴   Point of intersection : P (-2, -1)    
         From (1) slope: (x)' y + (y)' x = 0

                
               Let 'θ' be the angle between the curves then     

                  
4. Show that the curves    and    cut each other at (2, 1) at an angle of  
Sol:           Given point :    P (2, 1)
                   
   

             
         Let  'θ'  be the angle between the curves then,
          
                               ⇒  tan θ = 1
                             ∴  θ     =  

5.   Show that the curves y² = 4 (x+1) and y² = 36 (9 - x) intersect orthogonally.
Sol:   Given curves: y² = 4 (x + 1)   ....  (1)
                                   y² = 36 (9 - x)   ....  (2)
                              From (1) and (2)
                               4(x + 1) = 36 (9 - x)
                               x + 1 = 81 - 9x
                               10x = 80
                             ∴    x = 8
                       From (1) : y² = 4 (8 + 1)
                        ⇒     y = 6
                      Point of intersection : (8, 6)
                    

                                    Clearly m₁ m₂ = -1
                      ∴  The given two curves cut each other orthogonally.

6. If the curves x = y² and xy = k intersect each other orthogonally then prove that 8k² = 1
Sol:      Given curves : x = y²   ....   (1)    
                                      xy = k  ..... (2)        From (1) and (2)
                                             y³ = k
                                        
      
                            From (2) :  Slope   :  (x)' y + (y)' x = 0 

                     Since given two curves cut each other orthogonally
                                  ⇒  m₁ m₂ = -1 
                        
                                  
                                  Cubing on both sides
                                 ∴           8k² = 1

7. If the curves y² = 4ax and xy = c² intersect each other orthogonally then prove that c⁴ = 32 a⁴
Sol :      Given curves :  y²  =  4ax .... (1)
                                         xy  =  c²  .... (2)
                      Let the point of intersection be (x, y)
              
       Since given two curves cut each other orthogonally
                                                ⇒   m₁ m₂  =  -1
                                               
                                         ⇒   x = 2a
                      From (1) : y² = 4a (2a)
                           ⇒  y² = 8 a²
                          ⇒   y = 2 a
              ∴ Point of intersection: (2a, 2 a)
                       From (2) : c² = xy
                          ⇒   c² = (2a) (2  a)
                        ⇒     c² = 4  a2
                      Squaring on both sides
                     ∴       c⁴ = 32 a⁴

8.   Show that the curves y = x³ - 3 x² - 8 x - 4 and y = 3x² + 7 x + 4 touch each other at (-1, 0).  Also find the equation of the common tangent and common normal at (-1, 0).
Sol:    Given curves   y = x³ - 3 x² - 8 x - 4 ..... (1)
                                    y = 3 x² + 7 x + 4 .... (2)
                             Given point P (-1, 0)
                          
    
                                       ∴    m₁ =  m₂
         The given two curves touch each other at (-1, 0)
          Equation of the common tangent : y - 0 = 1 (x + 1)
                                                                  ⇒   x - y + 1 = 0.
          Equation of the common normal : y - 0 = -1 (x + 1)
                                                                ⇒      x + y + 1 = 0

9.  At any point on the curve y² = 4ax, a > 0 show that the length of the normal is constant
and the length of the sub tangent is twice the x - coordinate of the point.
Sol:            Given curve y² = 4ax
                     Let the point be P (at², 2at)
                    
                                 
                    
                                           = 2a which is constant
                        
                                 2at²        =  2 [X - coordinate of the point]

1. Find the points of local maximum (or) local minimum for the function x³ - 9x² + 24x - 12
Sol:      Let f(x) = x³ - 9x² + 24x - 12
                ⇒   f' (x) = 3x² - 18x + 24
               ⇒    f" (x) = 6x - 18
            For maximum (or) minimum f' (x) = 0
             ⇒      3x² - 18x + 24 = 0
             ⇒   x² - 6x + 8 = 0
             ⇒     x² - 2x - 4x + 8 = 0
             ⇒   x (x - 2) - 4 ( x - 2) = 0
             ⇒     (x - 2) (x - 4) = 0
                    ∴    x = 2 ;  x = 4
            At x = 2;   f"(2)  =  6(2) - 18 = - 6 < 0
            At x = 2,  f(x) has maximum value
∴ Local maximum value f(2) =    8 - 36 + 48 - 12 
                                                ⇒    f(2) = 8
                       At x = 4: f"(4) = 24 - 18 = 6 > 0
                       At x = 4, f(x) has minimum value
∴ Local minimum value f(4) =    64 - 144 + 96 - 12
                                                   f(4) = 4
2.   Show that the function     has a minimum value at x = e
sol:    Given  :      
                                             
                 
             for minimum value, f' (x)  = 0    
                                                     
                                                    ⇒      log x = 1
                                            ∴   x = e   ;     At x = e;
                                                    

                                          ⇒    f" (e) = 1/e > 0
                           ∴   At x = e, f(x) has minimum value.
 

3.  Show that y = sin³ x cos x has a maximum value at x =   and find its value.
Sol :    Given: y = sin³ x cos x
                       
                        
                             for maximum value    =  0
                          ⇒   3 sin² x cos² x - sin⁴x = 0
                          ⇒  sin² x (3cos2x - sin2 x) = 0
                          ⇒   sin²x = 0 ;   ⇒ 3 cos²x - sin²x = 0
                          ⇒ sin x = 0  ;    ⇒ 3 cos²x = sin²x
                            ∴   x = 0       ;   ⇒  tan²x = 3
                                     ⇒    tan x =  
                                         ∴   x =   
                                At x =    ;
                           
   
            ∴ At x =   , y = sin3x cos x has maximum value.
                   ∴  Maximum value: sin³  cos 
                                              
                                                       
4.  Show that the semi-vertical angle of the cone of maximum volume and of given slant height is tan-1()
Sol :  Let '' be the semi-vertical angle of the cone.

        Let r, h, l be the radius, height and slant height of the cone respectively.
                         Volume of the cone: V =  π r²h 

                                                                                         
          

      
                 For maximum value of V :    = 0     
                       ⇒   2 sin α  cos²α - sin³ α  = 0
                        ⇒  2 cos² α  - sin² α   = 0
                       ⇒   2 cos² α   = sin² α
                            =  tan² α   =  2
                      ⇒   tan α =  
                       ∴   α =  tan-1 ()
            ∴ If V is maximum then semi-vertical angle is tan-1 ().

5.   Show that the maximum rectangle inscribed in a circle is a square.
Sol:      Let 'l' be the length and 'b' be the breadth of the rectangle.
             Let ABCD be the rectangle inscribed in a circle of radius 'r'.
                          ∴  l² + b² = (2r)²
                                            

                          
                     For maximum value :     = 0
                                                            
                                                               4 r² - 2 l² = 0
                                                               2 l² = 4r²
                                                               l² = 2r²
                                                            ∴   l =  r
                               
                                                     
                
                                    ∴  A is maximum when l = b
                     Hence the maximum rectangle inscribed in a circle is a square.
 

6. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius 'a' is   
Sol: Let R be the radius and H be the height of the cylinder.
          From  ∆OAB, OA²  +  AB² =  OB²
               
                                         
               Volume of the cylinder : V =  πR² H
                                                                     
                   For maximum value :     =  0
           
 

7. A cylinder is inscribed in a sphere of radius R. Show that the height of the cylinder is R when its surface area is maximum.
Sol:           Let 'O' be the centre of the sphere of radius R.
            Let 'r' be the radius and 'h' be the height of the cylinder inscribed in a sphere.

              
             
                           
                                              
  For maximum (or) minimum :     = 0
                 
           
              ⇒      h² = 2R²
                      ∴     h =  R
       Hence the surface area of the cylinder is maximum when h =  R

Writer Sayyad Anwar

1.  A conical vessel of height 10 ft and semi vertical angle 30⁰ is full of water. It empties in such a way that the height of the water in the vessel is decreasing at a constant rate of one inch per minute. Find the rate at which the volume of the water in the vessel is decreasing when its height is 6 feet.
Sol: Let r be the radius and h be the height of the cone.


 

2. A conical vessel whose vertical angle is 90o is placed with its axis vertical and the vertix down wards. If water flows into it at the rate of 1 c.ft. per minute, find the rate of which the level of water is rising when the height of the water in it is 2 ft.
Sol: Let 'r' be the readings and 'h' be the height of the conical vessel.
Given: vertical angle: 90^o
   semi-vertical angle:  α  =  45^o

3. a particle is moving on a straight line and the distance 's' described by the particle in time 't' is given by     Find when its velocity vanishes what is the maximum velocity?

4. Aconical vessel whose height is 1 mt. and the radius of whose base is 5 mt. is being filled with water at the rate of   cubic meter per minute. Find the rate at which the level of the water in the vessel is rising, when the depth is 4 meters.
Sol: Let 'r' be the radius and 'h' be the height of the conical vessel.
Given h  =  2r

 

5. The base radius of a cylindrical vessel full of oil is 30 cms. Oil is drawn at the rate of 27,000 c. cm per minute. Find the rate at which the level of the oil is falling in the vessel ?
Sol : Let r, h, v be the radius, height, volume of a cylindrical vessel respectively.

 

6. A balloon which is always remains spherical, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon is increasing
when the radius is 15 cm.
Sol: Let r and v be the dimensions of the spherical balloon.

 

7. The surface area of a sphere increases at the rate of 20 Sq.cm per sec. At what rate does the volume of the sphere increases, when the volume is 36 π c.c ?
Sol: Let 'r' be the radius, A be the surface area, v be the volume of the sphere.


 

8. The radius of a circular disc increases at a uniform rate of 0.025 cm. per second. Find the rate at which the area of the disc increases when its radius is 15 cm.
Sol: Let r be the radius and A be the area of a circular disc.

9. A balloon is in the shape of an inverted cone surmounted by hemisphere. Diameter of the sphere is equal to the height of the cone. If h is the total height of the balloon, then how does the volume of the balloon changes with h? what is the rate of change in volume when h = 9 unit ?
Sol: Let r be the radius and v be the volume of the hemisphere.
Let r, h, v be the radius, height and volume of the cone.
Given: height of the balloon h  =  3r
and height of the cone : 2r


 

10. A man 180 cm high, walks at a uniform rate of 12 km per hour away from a lamp post of 450 cm high. Find the rate at which the length of his shadow increases.
Sol: AB is lamp post and PQ is person x = distance of the person from the lamp post
y = length of the shadow

Writer Sayyad Anwar

 

1.  If y = x² + 3x + 6,  x = 10,  δx = 0.01 then find  y and df

Sol:  Given y = x² + 3x + 6

                    x =  10

                 δx =  0.01

          

         ⇒  δy =  f(10 + 0.01) - f(10)

         ⇒   δy =  f(10.01) - f(10)

         ⇒  δy =  [(10.01)2 + 3 (10.01) + 6]  -  [(10)2 + 3(10) + 6]

         ⇒  δy = 100.2001 + 30.03 + 6 - 100 - 30 - 6

         ⇒    δy = 0.2001 + 0.03

            ∴  δy = 0.2301

               

            ⇒  dy = (2x + 3) x

            ⇒  dy = [2(10) + 3] (0.01)

            ∴ dy  =  0.23

2. If y = log x, x = 3,  δx = 0.003 then find  y and dy

Sol.     Given y = log x

                       x = 3

                     δx = 0.003

              

           ⇒    δy = f (3 + 0.003) - f (3)

               δy = f (3.003) - f (3)

               δy = log (3.003) - log 3     

               δy = log ()

             ∴   δy = log (1.001)

      

  

3.  Find the approximate value of  

Sol:    Let:       f(x + δx)  =  

                                     δx  =  81

                                    δx = 1

                                  f(x) =  

           

4.   Find the approximate value of  

Sol:     Let:   f(x + δx) =   

                                  x  =  64

                               δx  =  - 1

                              f(x)  =  

        

        
         

5. Find the approximate value of 

Sol:     Let :    f(x +δx) =  

                                     x = 25

                                   δx = 0.2

                                 f(x) = 

              

        

6.   Find the approximate value of  

sol:     Let:   f (x + δx) = 

                      x = 8  ;        δ x = -0.2

                                f(x) = 

              

            

7.   Find the approximate value of sin 60º 1'  (Given that 1' = 0.1745 radian).

sol.     Let:     f (x + δx) = sin 60º 1'

                                    x = 60º

                                

                                f(x) = sin x   

                 

            

8.   Find the approximate value of cos  45º 6'.  (Given that 1' = 0.1745 radian).

Sol.      Let:   f (x + δx)  =  cos 45º 6'

                                     x = 45º

                                 

                                 f(x) =  cosx

                                

                         

9. If the increase in the side of a square is 1% then find the percentage of change in the area of  the square.

Sol:    Let  'x' be the side and 'A' be the area of square.

Area of square: A = x²  

                             

                            

10. The diameter of a sphere is measured to be 20 cm. If an error of 0.02 cm occurs in this, find the errors in volume and surface area of the sphere.

Sol:   Let 'd' be the diameter, 'V' be the volume and 'S' be the surface area of sphere.

          Given: d = 20  ;    δd = 0.02  ;    δV,   δS = ?

        Volume of sphere:

                     
                                 

      ∴  Error in volume :   δV : 4 πcm³

     Surface area of sphere:  S =   πd²

                                                  

                                       ⇒   δS = 2πd . δd

                                       ⇒   δS = 2π (20) (0.02)

      ∴  Error in surface area:  δS = 0.8π sq.cm.

11. Area of a  ∆ABC is measured by the measures of a, b, c. If c is the error in measuring c, then what is the percentage error in the area.

sol:     Let 's' be the area of triangle.

      Given:   S  =   ab  sinC                                                  

         

        ⇒   δs  =   ab cos C δc          

    Percentage error in Area:

                                      

                                    
                             = 100 cotC . δc

12. The time of a complete oscillation of a simple pendulum of length l is given by the equation  where g is gravitational constant. Find the approximate error in the calculated g corresponding to an error of 0.01 percent in the value of t.

Sol:           Given:

                                             

                Given :

                                                         

                             Taking " log" on both sides

                

        ∴    Percentage error in g : - 0.02.

 

 

Writer Sayyad Anwar 

Important Questions

 

4. Find   if x = 3 cos t - 2 cos³ t,  y = 3 sin t - 2 sin3 t.
Sol:   = -3 sin t + 6 cos² t sin t
               =  3 sin t (-1 + 2 cos2 t)
               =  3 sin t cos ²t.
            =  3 cos t - 6 sin² t cos t
               =  3 cos t (1-2 sin² t)
               =  3 cos t cos 2t
      
             =  cot t.

5. Find the derivative of the function x = log (1+ sin²y) w.r.t. x.
Sol:

Alternate Method
                                      x = log (1 + sin²y) 
                                      Differentiate w.r.t. y 

 

Short Answer Type Questions                                4 Marks
 

1. Find the derivative of x sin x w.r.t. x from first principles.
Sol: Let f(x) = x sin x

 

2. If sin y = x sin (a+ y) then show that   , (a is not a multiple of π)
Sol: sin y = x sin (a + y) 
       
     Diff. w.r.t. 'x' on both sides

Sol: x^y = e ^x-y
        taking log on both sides
                   y log x = x - y
            

4. Find  the derivative of sin^-1
Sol: Let y =  sin^-1
       Let 2^x = tan θ

5. If x = a (t - sin t), y = a (1 + cos t) find 
Sol:    =  a (1 - cos t)                            =  a (0 - sin t)

6. If ax² + 2hxy + by² = 1 then prove that  
Sol: ax² + 2hxy + by² = 1         Diff. x on both sides

                  
                             (using (1)
  

7 Marks Long Answer type Question

1. Find the derivative of    a, b > 0
Sol: Let y = 

2. If   then Show that  
Sol: 
             Diff. w.r.t. 'x' on both sides

3. If   then 
Sol:
 
   Let x =  tan θ 

4. Find the derivative of 
Sol: Let U = (sinx) ^logx     and V = x ^sinx
              Taking log on both sides 
         log U = log x. log (sinx)
             diff. w.r.t. x on both sides

Taking log on both sides
        log V = sin x. log x
diff. w.r.t. 'x' on both sides

Now   y = U + V

5. If x^y + y^x =  a^b then show that   
Sol: Let U = x^y  and V = y^x
           Taking log on both sides
          log U = y logx
            Diff. w.r.t. 'x' on both sides

         
Taking log on both sides
              log V = x logy
Diff. w.r.t. 'x' on both sides

Now, xy + yx = a^b  U+V = a^b

 

Writer S. V. Sailaja

 

Writer C. Sadasiva Sastry

1. Find the angle between the two planes x  +  2y  +  2z - 5  =  0 and  3x  + 3y  +  2z - 8  = 0

Sol: Given Planes: x  +  2y  +  2z  -  5 =  0  .........( 1 )

3x  + 3y  + 2z  - 8  =  0  ........( 2 )
                   
                    
                  

2. Find the Perpendicular distance of the plane 2x + y + 2z - 17 =  0 from the point (2, 1, 0)

Sol: Given plane  2x + y + 2z  - 17 =  0

Given Point  ( 2,  1,  0 )

The Perpendicular distance from ( x1,  y1,  z1 ) to the plane ax  +  by  +  cz  +  d  =  0  is

3. Find the Equation fo the plane through  ( - 2,  - 1,  - 3 ) and parallel to the plane 2x  +  y  +  z   - 9   =  0

Sol: Given Plane 2x + y + z - 9  =  0

         Given point P ( - 2,  - 1,  - 3 )

Equation of the plane through ( x₁,  y₁,  z₁ ) and parallel to ax + by + cz + d  =  0 is

a ( x - x₁ )  +  b ( y - y₁ )  +  c ( z -z₁ )  =  0

Required plane is  2 ( x + 2 )  +  1 ( y + 1 )  +  1  ( z + 3 )  =  0 

         =  2x  +  y  +  z  +  8  =  0

 ∴   Required plane is 2x  +  y  +  z  +  8  =  0

4. Find the Equation of the plane through ( - 1,  0,  - 6 ) and perpendicular to the line whose direction ratios are ( 6,  20,  - 1 )

Sol: Equation of the plane through ( x₁,  y₁,  z₁ ) and perpendicular to the line whose

direction ratios are ( a,  b,  c  ) is a ( x - x₁ )  +  b  ( y -y₁ )  +  c (  z - z₁ )  =  0

Required plane is  6 ( x + 1 )  + 20  ( y - 0 )  -1  ( z + 6 )  =  0 

                =  6x  +  20y - z  =  0

 ∴  Required plane is  6x  +  20y - z  =  0

5. Find the distance between the parallel planes 2x  +  y - 2z  +  8  =  0 and   2x + y -2z - 19  =  0

Sol: Given parallel planes 2x  +  y  -  2z  +  8  =  0 ......... ( 1 )

                                             2x + y - 2z - 19  =  0 ......... ( 2 )

Distance between the parallel planes ax + by + cz + d₁ = 0,  ax + by + cz + d₂  =  0  is

6. Find the ratio in which the plane 2x + 3y - 2z + y  =  0 divides the line joining the points ( - 1,  2 , 3 ) and ( 2,  3,  5 )

Sol: Given plane 2x + 3y - 2z + 7  =  0

Given points (- 1,  2,  3 ) and  ( 2,  3,  5 )

The ratio in which the plane  π  ≡  ax + by + cz + d  =  0 divides the line segment joint

 A  ( x₁,  y₁,  z₁ ) and B ( x₂,  y₂,  z₂ )

is  –π₁₁₁  :  π₂₂  where  π₁₁₁  =  ax₁ + by₁ + cz₁ + d, and  π₂₂₂  =  ax₂ + by₂ + cz₂+ d

Required ratio =  - π₁₁₁  :  π₂₂₂
          ⇒     - [ 2 ( -1 )  +  3 ( 2 )  -  2 ( 3 )  +  7  ]   :   [  2 ( 2 )  +  3 ( 3 )  -  2 ( 5 )  +  7  ]
         ⇒       - 2 :10
           =   1  :  5   Externally
∴    The plane divides the points externally in  1  :  5  ratio

7. Show that the points ( 2,  3,  4 ),  ( 1,  2,  3 )  lie  in  the  same  side  of  the  plane 3x  -  2y  +  z  -  5   =   0

Sol: Given plane 3x - 2y + z - 5  =  0
Given points A  ( 2,  3,  4 ),  B  ( 1,  2,  3 )
The points A and B lie in the same side (or) Opposite side of the plane π  =  0
According as π₁₁₁ , π₂₂₂ have the same sign (or) oppositic signs.
⇒   π ₁₁₁     =    3 ( 2 ) - 2 ( 3 )  +  4  - 5
⇒   π ₁₁₁     =    6  -  6  +  4  - 5
⇒   π ₁₁₁     =    - 1  <  0
⇒   π ₂₂₂    =    3 ( 1 )  - 2 ( 2 )  + 3  - 5
⇒      π ₂₂₂    =    3 - 4  +  3  - 5
⇒     π ₂₂₂    =    - 3  <  0 
∴     π 111 and π 222 both have same sign given points lie in the same side of plane.

8. Find the equation of the plane having intercepts -4, 5, 6 on the x, y and z axes respectively.

Sol: Given  x  -  intercept :  a  =  -  4
                     y  -   intercopt :  b  =     5
                     z  -    intercopt :  c  =     6
Equation of the plane having x -  intercept  a, y -  intercept  b, z -  intercept  c  is

9. Show that  ( 2,  -3,  1 ) lies in the origin side of  2x + 3y + 4z + 7  =  0

Sol: Given plane  2x + 3y + 4z + 7  =  0
Given point ( 2,  - 3,  1 )

10. Write the equation of the plane 5x + 2y - 3z - 30  =  0 in the intercept form

Sol:  Given Plane 5x + 2y – 3z – 30  =  0
                   5x + 2y – 3z  =  30

11. Reduce the eqation  2x - 2y + z + 6  =  0 of a plane into normal form.

Sol: Given plane  2x - 2y + z + 6 =  0

Writer  -  Anwar

1. A Line makes angles  and  with the positive axes of x and y respectively. Find the angle that it makes with the positive axis of z.
Sol:                              Given: α=     β = 
                                       Required Angle: γ

2. If (3,-4,12) are direction ratios of a line then find the direction cosines of the line.
Sol:                      
                                Given:   a = 3 | b = -4|,    c = 12

3. Find the direction ratios and direction cosines of the line joining the points (5,-2,3) and 
       (-2,3,7)
Sol:                          Given Points: A =  (5,-2,3) 

                                    B =  (-2,3,7)

4.  Find the angle between the lines whose direction ratios are (1, −2, 1) and (−1, 1, 0)
Sol: 
                   Given d.r.'s: (a₁,b₁,c₁) =  (1,−2,1)                (a₂,b₂,c₂) = (−1,1,0)

          
                                          θ  =  150°
 Angles between the given lines  are  30°, 150°

5. Find the equations of the line passing through (2,−3,1) and having direction ratios 
       (3,2,−5)
sol:

6. Find the equations of the line passing through (−2, 3, −1) and (3, 4, 2)
Sol:   
                            Given Points:  (x₁, y₁,  z₁)  =  (−2, 3, −1)

                                                     (x₂,  y₂,  z₂)  =  (3, 4, 2)

 

7. Find the coordinates of the foot of the perpendicular drawn from A(1,0,3) to the line joining the points B (4,7,1) and C (3,5,3)
Sol:   Foot of the perpendicular means the points of intersection of two perpendicular lines.
Let foot of the perpendicular :  D
Let D divide   in the ratio λ :  1

                              ⇒  2λ  + 3 + 10λ + 14 + 4  =  0

                             ⇒   12λ  +  21  = 0

∴ Foot of the Perpendicular :

⇒ 
⇒ 
                                                     
8.   Find the length of the projection of the join of P (3, 4, 5) and Q( 4, 6, 3) on the line joining  A (−1, 2,4 ) and  B (1, 0, 5)
sol:  Given Points:   A (−1, 2, 4),    B (1, 0, 5)
                                   P (3, 4, 5),      Q (4, 6, 3)
         D.r.s of  : (1+1,  0−2,  5−4)
                  ⇒    (2,−2,1)
         
                     ⇒            
                   
 ∴  Projection of the lin e segment join of P (3, 4, 5) and Q (4, 6, 3) on
     : =  |  l (X₂ - X₁) + m (y₂ - y₁) + n (z₂ - z₁)  |

9.  Find the angle between the lines whose d.c.s are given by the equations 3l +  m +  5n = 0
      and 6mn − 2nl  +  5lm = 0
Sol:                      Given Lines: 3l  +  m  +  5n = 0 −−−  (1)
                          ⇒    m =  −  (3l +  5n)
                              6mn −  2nl  +  5lm = 0 −−−  (2)
                                from  (1)  and  (2)
                            −6n(3l + 5n) − 2nl − 5l(3l + 5n) = 0
                      ⇒       − 18nl  −  30n²  −  2nl − 15l² − 25nl = 0
                        ⇒     −15l ² − 45nl − 30n² =0
                        ⇒     l2 + 3nl + 2n² = 0
                             l² + 2nl + nl + 2n² = 0

                          ⇒   l(l + 2n) + n(l + 2n) = 0
                          ⇒   (l + n)(l + 2n) = 0
                           ⇒  l  +  n = 0 −−−  (3)    l + 2n = 0 −−−  (4)
                                Solving  (1)  and  (3)
                                3l +  m +  5n = 0
                                l + 0.m +  n = 0
                      
                            D.r.s of one Line :  (1, 2, −1)

                                Solving (1) and (4)
                      
                                                       
                            D.rs of Second Line: (2, −1, −1)
                   Let 'θ' be the angle between the lines then

                     

10.  If a line makes angles α,  β,  γ,  δ with the four diagonals of a Cube then prove that  cos ² α+ cos² β + cos² γ + cos² δ =   
Sol:   
                                                                                                  

Let the length of an edge of the Cube be 1
Let one corner of the cube be taken as the origin and the three edges through this corner as
Direction Cosines and Direction Ratios.. (New Syllabus)  >> Page - 14

the axes from the figure,
                                                   A  =  (1,  0,  0)
                                                   B  =  (0,  1,  0)
                                                   C  =  (0,  0,  1)
                                                   P  =  (1,  1,  1)
                                                   Q  =  (1,  1,  0)
                                                   R  =  (0,  1,  1)
                                                   S  =  (1,  0,  1)
                        
                        
Let (l, m, n) be the D.c.s of the given line and α, β, γ, δ be its inclinations with the diagonals.
Then
                  
                 

                     cos² α  +  cos² β  +  cos² γ  +  cos² δ  =

             [ (l + m + n) ²  +  ( -l + m + n ) ²  +  ( l - m + n ) ²  +  ( l + m - n) ² ]
 
                              =     [ l 2 + m2 + n2 ]

                              =     (1)                           [ ... l 2 + m2 + n2 = 1]
                              =   

               cos² α + cos² β + cos² γ + cos² δ =  

                                  
                                 

1. Find the distance between the points (2, 3, 1) and (-3, 1, -5)

Sol : Given Points:  A  =  ( 2, 3, 1 )

                                    B  =  ( -3,  1, -5 )

2. Show that the points A ( 2, -1,  3 ), B ( 4, -2, 1 ), C ( 4, 5, -7) and D ( 2, 6, -5 ) form a parallelogram

Sol: Given Points

A    =   ( 2, -1, 3 )

B    =   ( 4, -2, 1 )

C    =   ( 4,  5, -7 )

D    =  ( 2,  6,  -5 )

3. Show that the points A(0, 1, 2), B(2, -1, 3) and C(1, -3, 1) form a right angled isosceles triangle

Sol : Given points;  A   =  ( 0,  1,  2 )

                                    B  =  ( 2,  -1,  3)

4. Show that the points A ( 3,  -2,  4), B(1,  1,  1) and C(-1,  4,  -2) are Collinear

sol: Given Points A  =  (3,  -2,  4)

                                B  =  ( 1,  1,  1)

                                C  =  (-1,  4, -2)

5. Show that the four points A (-2,  4,  1) B (-1,  5,  5), C (2,  2,  5) and D  (1,  1,  1) form a Square

Sol : Given points A  =  (-2,  4,  1), B  =  (-1,  5,  5), C  =  (2,  2,  5), D  =  (1,  1,  1)

6. Find the coordinates of the point which divides the join of A(2,  -1,  4) and B(4,  3,  2) in the ratio 2 : 3
Sol : Given points A  =  (2,  -1,  4) B  =  (4,  3,  2)
Given ratio m  :  n  =  2 : 3

7. Find the ratio in which the line joining the points (1,  2,  3) and (-3,  4,  -5) is divided by the
xy - plane.
Sol: Given points (1,  2,  3) and (-3,  4,  -5)

Required Ratio - z₁  :   z₂
   - 3   :  - 5
     3    :    5

8. Find the Centroid of the triangle formed by the points (7,  - 4,  7), (1,  - 6,  10), (5  - 1,  1)
Sol : Given points A  =  (7, - 4,  7) B = (1, - 6,  10) C = (5, - 1,  1)

9.Find the centroid of the tetrahedron formed by the points (2, 3, -4), (-3, 3, -2), (-1, 4, 2), (3, 5, 1)
Sol: Given Points
A  =  (2,  3,  -4)
B  =  (-3,  3,  -2)
C  =  (-1,  4,  2)
D  =  (3,  5,  1)

10. If (2, 4, -1), (3, 6, -1), (4, 5, 1) are the consecutive vertices of a parallelogram respectively, find its fourth vertex.
Sol. Given Three Vertices A = (2, 4, -1)
B = (3, 6, -1)
C = (4, 5, 1)
Let the fourth vertex be : D
  
     D  =  ( 2  +  4  - 3,  4  +  5  -  6,  - 1  +  1  +  1 )
   Fourth Vertex: D  =  ( 3,  3,  1 )

11. If (3,  2,  -1), (4,  1,  1), (6,  2,  5) are three vertices and (4,  2,  2) is the centroid of a tetrahedron, find the fourth vertex of the tetrahedron.
Sol: Given Vertices: (3,  2,  -1), (4,  1,  1), (6,  2,  5)
         Given Centroid: (4,  2,  2)
   Let the fourth vertex be :  ( x₄,  y₄,  z₄ )

        
       

1. If scope of one line is double the scope of other line of the pair of lines ax² + 2hxy + 6y² = 0 then show that 8h² = 9ab.
Sol: Given equation is ax² + 2hxy + 6y² = 0
        If m1, m2 are the scopes of the pair of lines, then we have
                  
          But given that, m₂ = 2m₁

2. If 2x² + kxy - 6y² + 3x + y + 1 = 0 represents pair of lines, find k.
Sol: Given equation is   2x2 + kxy - 6y2 + 3x + y + 1 = 0
                          a = 2,     2h = k,     b = -6,   2g = 3,    2f = 1,    c = 1
        Since the given equation represents pair of lines, we have   = 0
            i.e.    abc + 2fgh - af² - bg² - ch² = 0
         

3. Prove that the two pairs of lines 3x² + 8xy - 3y² = 0 and 3x² + 8xy - 3y² + 2x - 4y - 1 = 0 form a square.
Sol: Given pairs are
                                    3x² + 8xy - 3y² = 0 ............... (1)
                            and 3x² + 8xy - 3y² + 2x - 4y - 1 = 0 .................. (2)
Since coefficient of x² + coefficient y² = 3 - 3
                                                                  = 0, in both the pairs,
each pair represents 2 perpendicular lines.
Since 1st equation is homogeneous, it represents pair of perpendicular lines passing through origin.
And 2^nd pair also represents a pair of perpendicular lines but not passing through origin.
Since the 2^nd degree terms of both the equations are same, the lines of one pair are parallel to the lines of other pair and hence they form a parallelogram with all the angles at vertices as 90º . Hence they form a rectangle. Now we shall verify whether diagonals are also at right angles, to confirm that the formed rectangle is a square.
Now (2) - (1) gives equation of one of the diagonals, that does not pass through origin.
And it is 2x - 4y - 1 = 0

  Scope of this diagonal m₁ = -2 /- 4    =  1 / 2
Now if C is the point of intersection of 2nd pair,
           
       Where a = 3,   2h = 8,   b = -3,   2g = 2,   2f = -4,   c = -1
               a = 3,   h = 4,   b = -3,   g = 1,   f = -2,   c = -1

 

4. Find the distance between the pair of parallel lines 9x² - 6xy + y² + 18x - 6y + 8 = 0
Sol:  Given equation is  9x² - 6xy + y² + 18x - 6y + 8 = 0
         Here    a = 9,   2h = - 6,     b = 1,     2g = 18,     2f  = - 6,     c = 8
               i.e. a = 9,       h = -3,      b = 1,       g = 9,         f = -3,      c = 8
As h² = ab, the given pair of lines are parallel and, distance between them is given by

 

5. Find the angle between the pair of lines obtained by joining the origin to the points of intersection of x2 + 2xy + y2 + 2x + 2y - 5 = 0. And the line 3x - y + 1 = 0.
Sol: Given curve is x² + 2xy + y² + 2x + 2y - 5 = 0 ............. (1)
       Given line is 3x - y + 1 = 0
                        i.e. y - 3x = 1 ............. (2)
Now we obtain the required pair of lines by homogenising eqn (1) using eqn (2).
          We have x2 + 2xy + y2 + (2x + 2y) × 1 - 5 × 12 = 0
        i.e.  x² + 2xy + y² + 2(x + y) (y - 3x) - 5 × (y - 3x)2 = 0
        i.e.  x² + 2xy + y² + 2{xy - 3x2 + y² - 3xy} - 5 {y² - 6xy + 9x²} = 0
        i.e.  x² + 2xy + y² + 2xy - 6x² + 2y² - 6xy - 5y² + 30xy - 45 x² = 0
        i.e. -50 x² + 28xy - 2y² = 0
        i.e.   25 x² - 14xy + y² = 0


 
6. Show that the pair of lines obtained by joining the origin to the point of intersection of 6x - y + 8 = 0 and 3x² + 4xy - 4y² - 11x + 2y + 6 = 0 are equally inclined which the co-ordinate axes.
sol: Given curve 3x² + 4xy - 4y² - 11x + 2y + 6 = 0   ............ (1)
       Given line 6x - y + 8 = 0
                          i.e. y - 6x = 8
                   
   Now to obtain the required pair of lines, we have to homogenise (1) using (2).
  Thus we have,     3x² + 4xy - 4y² + (-11x + 2y) × 1 + 6 × 12 = 0

Now angle bisectors of this pair is given by h(x² - y²) - (a - b)xy = 0 where a = 4
                                  b = -1
                                 2h = 0
                                0 - (4 + 1)xy = 0
                                       xy = 0
                                  i.e. x = 0    or   y = 0,
Which are coordinate axes.
i.e.    The pair of lines formed by going the origin to the points of intersection of given curve and line, are equally inclined whith the co-ordinate axes.
 

Writer Karusala Anjan

1. Find the equation of the straight line passing through the point (x1,  y1) and making X and Y intercepts which are in the ratio m : n.

Sol: Given that X and Y intercepts are in the ratio m : n

           The intercepts may be taken as ma and na  such that  ma : na = m : n.

           The line in intercept form takes the equation =  

                                                                    (or)   nx + my = mna  ----------→ (1)

        If this line passes through (x₁, y₁), we have nx₁ + my₁ = mna

                  (1)   nx + my = nx₁ + my₁ is the required equation.

2. A(10, 4), B(−4, 9), C(−2, −1) are the vertices of a triangle. Find the  equations of

     (i)                           (ii) The median through A       (iii) The altitude through B

     (iv) The perpendicular bisector of the side  .

Sol: (i) A = (10, 4), B = (−4, 9)

                The equation of AB =    
                                                       

                         5(x − 10) = −14(y − 4)    5x + 14y − 50 − 56 = 0

                         5x + 14y − 106 = 0

(ii) The median through A bisects the side BC i.e., passes through the  mid point of BC.

         B = (−4, 9), C = (−2, −1) implies mid point of BC = (−3, 4) = D  (say) and A = (10, 4).

           Equation of AD is y = 4  (or)  y - 4 = 0.

This is the eq. of the median through A.

(iii) The altitude through B is perpendicular to the side AC.

           A = (10, 4), C = (−2, −1) 

         Slope AC = 

           The slope of the altitude is  and the altitude passes through B = (−4, 9)

           Equation of the altitude through B is y − y₁ = m(x − x₁)

                     y − 9 =   (x + 4)

                    5y − 45 = −12x − 48

                    12x + 5y + 3 = 0

(iv) A = (10, 4), B = (−4, 9)

          

       Slope of its perpendicular =   

                           Mid point of AB = 

  Perpendicular bisector of AB passes through    and with slope  .

  Equation is y−y₁ = m(x−x₁)

                     y−  =  (x−3)

                     5y−  = 14x − 42

                     10y − 65 = 28x − 84    28x − 10y − 19 = 0.

3. If the portion of a line intercepted between the coordinate axes is cut at a point in the ratio m : n, then find the equation of the line.

Sol: Let   (or) bx + ay = ab be the line in intercept form.

 'a' is the X − intercept,   A = (a, 0) is a point on the X − axis.

 'b' is the Y - intercept,    B = (0, b) is a point on the Y − axis.

This line AB is cut at (x1, y1) say in m : n ratio.

                           

Note: (i) If (x₁, y₁) is the mid point of AB, then m = n and hence, the  equation takes the form   

(ii) If the point of bisection is (2p, 2q), then the equation takes the  form  

4. If the linear equations ax + by + c = 0 (abc ≠ 0) and lx + my + n = 0  represent  the same line and   write the value of r in terms of m and b. 

Sol:  ax + by + c = 0 and lx + my + n = 0 represent the same line.

         

5. Transform the equation (2 + 5k)x − 3(1 + 2k)y + (2 − k) = 0 into L₁ + λL₂ = 0 and find the point of concurrence of the family of straight lines.

Sol: (2 + 5k)x − 3(1 + 2k)y + (2 − k) = 0 can be written as (2x − 3y + 2)+ k(5x − 6y − 1) = 0

           This is of the form L₁ + λL₂ = 0 where   L₁ ≡ 2x − 3y + 2 = 0, L₂ ≡ 5x − 6y − 1 = 0

           Solving these two equations, we get the point of concurrence.

                                     4x − 6y + 4 = 0

                                    5x − 6y − 1 = 0                                  

                                      x − 5   = 0

                                   x = 5, y = 4

                                                               Point = (5, 4).

6. Find the value of 'p', if the straight lines x + p = 0 , y + 2 = 0 and 3x + 2y +5 = 0 are concurrent.

Sol:    x + p = 0  x = −p

            y + 2 = 0   y = −2

            3x + 2y + 5 = 0

 3(−p) + 2(-2) + 5 = 0

                  −3p + 1 = 0 

                            p = 

7. Find the area of the triangle formed by the line 3x − 4y+ 12 = 0 with  the coordinate axes.

Sol:  3x − 4y + 12 = 0

          a = X − intercept =  = −4

           b = Y−intercept =  = 3

           Area =    square units.

8. If 3a + 2b + 4c = 0 then, show that the equation ax + by + c = 0  represents a family of concurrent straight lines and find the point of concurrency.

Sol:  3a+ 2b + 4c = 0  c = 

          ax + by + c = 0  ax + by 

           4ax + 4by − 3a − 2b = 0

           a(4x − 3) + b(4y − 2) = 0

           (4x − 3) +  (4y − 2) = 0

   For various values of a and b, this equation represents a set of  concurrent lines.

                           The point of Concurrence = 

9. Astraight line through P(3, 4) makes an angle of 60° with the positive  direction of X−axis. Find the coordinates of the points on the line which are 5 units  away from P.

Sol:  This problem is related to the straight line in symmetric form which is

           given as 

           here (x₁, y₁) = (3, 4); θ = 60°;   = 5

  
 

10. Find the angle between the lines y = 4 − 2x and y = 3x + 7.

Sol: m₁ = −2; m₂ = 3

11. Find the equation of the straight line  parallel to the line 2x + 3y +7 = 0 and passing through the point (5, 4).

Sol:  The equation of the straight line passing through (x1, y1) and  parallel to the line

          ax + by + c = 0 is a(x − x₁) + b(y − y₁) = 0.

          Here, a = 2, b = 3, x₁ = 5, y₁ = 4.

           The equation required:

                                                          2(x − 5) + 3(y − 4) = 0

 2x + 3y − 22 = 0
 

12. Find the equation of the line perpendicular to the line 5x − 3y + 1 =0 and passing through the point (4, −3).

Sol:  The equation of the line passing through (x₁, y₁) and perpendicular  to the line

          ax + by + c = 0 is b(x − x1) − a(y − y1) = 0.

          Here, a = 5; b = −3; x1 = 4; y1 = −3.

          The equation required:

                                                   −3(x − 4) − 5(y + 3 ) = 0

                                                             −3x − 5y − 3 = 0

                                                              3x + 5y + 3 = 0

13. Find the value of 'k', if the straight lines y − 3kx + 4 = 0 and (2k − 1)x − (8k − 1)y −6 = 0  are perpendicular.

Sol:  Slope of y − 3kx + 4 = 0  is  3k = m₁ (say)

          Slope of (2k − 1)x − (8k − 1)y − 6 = 0 is    = m2 (say)

          for perpendicular lines, m1m2 = −1

         

         6k²−3k = 1−8k (or) 6k² + 5k−1 = 0

         6k(k +1)−1 (k+1) = 0

          k = −1  or .

14. Find the equations of the straight lines passing through (1, 3) and (i) parallel to

(ii) perpendicular to the line passing through the points (3, −5) and (−6, 1).

Sol:  Slope of the line passing through (3, −5) and (−6, 1) = 

Equation of the line passing through (1, 3) and parallel to the line with slope =  isy − y₁ = m(x − x₁) y − y₁ = m(x − x₁) 

 y − 3 =   (x − 1)

  3y − 9 = −2x + 2

  2x + 3y − 11 = 0

Equation of the line passing through (1, 3) and perpendicular to the line with slope =  is

y − y₁ =   (x − x₁)

 y − 3 =   (x−1)

  2y − 6 = 3x − 3

  3x − 2y + 3 = 0.

15. The line   meets the X − axis at P. Find the equation of the line per pendicular to this line at P.

Sol:    represents the line in intercept form with X − intercept = a and Y − intercept = b in the 4th quadrant.

 P = (a, 0) and Q = (0, −b)

 

 16. Find the foot of the perpendicular drawn from (3, 0) upon the straight line 5x + 12y − 41 = 0.

Sol: If (h, k) is the foot of the perpendicular drawn from (x₁, y₁) upon the straight line ax + by + c = 0, then we have

17. Show that the distance of the point (6, −2) from the line 4x + 3y = 12 is half of the distance of the point (3, 4) from the line 4x − 3y = 12.

Sol:  The distance of the point (6, −2) from the line 4x + 3y − 12 = 0 is given by

The distance of the point (3, 4) from the line 4x−3y−12 = 0 is given by

 d₂ = d₁  the result proved.

18. Find the locus of the foot of the perpendicular from the origin to a variable straight line which always passes through a fixed point (a, b).

                                                       

Sol:  Let F(x₁, y₁) be a variable foot of the perpendicular drawn from origin that passes through P(a, b).

19. Find the equations of the straight lines passing through the point (−3, 2) and making an angle of 45° with the straight line 3x−y+4 = 0

Sol:  Let 'm1' be the slope of the required straight line that passes through the point (−3, 2).

 The equation of the line is y−2 = m₁(x + 3) → (1)

Let 'm₂' be the slope of the given line 3x−y + 4 = 0

 m₂ = 3

The required line makes an angle of 45° with the given line.

 ± (1 + 3m₁) = (m₁ − 3)²

 1 + 6m₁ + 9m₁₂ = m_{1 2} − 6m₁ + 9

 8m_{1 2} + 12m₁ − 8 = 0

 2m_{1 2} + 3m₁ − 2 = 0

 2m₁ (m₁ + 2) − 1(m₁ + 2) = 0

 (2m₁ − 1)(m₁ + 2) = 0

m₁  = 1/2 or −2.

Putting these values in (1), we get the equations as follows:

m1  =  1/2--- y−2 =  1/2(x + 3)

2y − 4 = x + 3

x − 2y + 7 = 0         If m1 = −2

y − 2 = −2(x + 3)

2x + y + 4 = 0

20. Find the equations of the straight lines passing through the point of intersection of the lines 3x + 2y + 4 = 0, 2x + 5y = 1 and whose distance from (2, −1) is 2.

Sol:  Let the line required be of the form L₁ + λL₂ = 0 where

L₁ ≡ 3x + 2y + 4 = 0; L₂ ≡ 2x + 5y − 1 = 0

 (3x + 2y + 4) + λ(2x + 5y − 1) = 0 → (1)

 (3 + 2λ)x + (2 + 5λ)y + (4 − λ) = 0

This has a distance '2' from (2, −1)

(4 − λ)² = 29λ² + 32λ + 13

28λ² + 40λ − 3 = 0

(2λ + 3)(14λ − 1) = 0

λ = 1/14 (or) λ = -3/2

λ = -3/2 , (1)   (3x + 2y + 4) +  (2x + 5y − 1) = 0

4x + 3y + 5 = 0

λ = -3/2, (1)  (3x + 2y + 4) −( -3/2)  (2x + 5y − 1) = 0

y − 1 = 0

21. Find the Orthocentre of the triangle formed by the lines 4x − 7y + 10 = 0, x + y = 5 and 7x + 4y = 15.

Sol: 4x − 7y + 10 = 0, 7x + 4y − 15 = 0 form a pair of mutually perpendicular lines.

Hence, the angle between them is a right angle. And the two lines intersect at the point (1, 2).

Therefore, the triangle is right angled at (1, 2) and hence, the Orthocentre is (1, 2).

22. Find the values of k, if the angle between the straight lines kx + y + 9 = 0 and

3x - y + 4 = 0 is

Sol:  a₁ = k;   a₂ = 3;

          b₁ = 1; b₂ = -1    

 5(k² + 1) = (3k - 1)²

 5k² + 5 = 9k² - 6k + 1

4k² - 6k - 4 = 0

 2k² - 3k - 2 = 0

 2k^{2 -} 4k + k - 2 = 0

 2k(k - 2) + 1(k - 2) = 0

 (2k + 1)(k - 2) = 0

 k = -1/2 or 2.

23. Find the equation of the straight line parallel to the line 3x + 4y = 7 and passing through the point of intersection of the lines x - 2y - 3 = 0 and x + 3y - 6 = 0.

Sol:  The required equation can take the form

L₁ + λL₂ = 0 which represents a line passing through the intersection of the lines L₁ = 0 and L₂  = 0 for various values of the parameter λ.

Here, we have

(x - 2y - 3) + λ(x + 3y - 6) = 0

i.e., (1 + λ)x + (3λ - 2)y - (3 + 6λ) = 0

This is parallel to 3x + 4y = 7 (given)

 Slopes are equal

 4 + 4λ = 9λ - 6

 10 = 5λ (or) λ = 2

 Line required: (x - 2y - 3) + 2(x + 3y - 6) = 0

 3x + 4y - 15 = 0

 

 

 

 

 

S. V. Sailaja.

1. Find the equation of locus of a point P such that PA² + PB² = 2C², where A = (a, 0),

     B = (-a, 0) and 0 < | a | < | c |

Sol: The given fixed points are A(a, 0), B(-a, 0)

let P(x, y) be a point on the locus.

The given geometric condition is PA² + PB² = 2C²

  (x-a)²+y²+(x+a)²+y² = 2C²

  y²+x²-2ax+a²+x²+2ax+a²+y² = 2C²

  2x²+2y²+2a² = 2C²

  x²+y² = c²-a²     0< | a | < | c |

Is the required equation of the locus.

2. Find the equation of locus of P, if the line segment joining A(2, 3) and B(-1, 5) subtends a right angle at P.

Sol: The given fixed points are A(2, 3) B(-1, 5)

Let P(x, y) be any on the locus, the geometric condition is AP²+BP² = AB²

  (x-2)²+(y-3)²+(x+1)²+(y-5)² = 32 + 22 = 13

  x²-4x+4+y²-6y+9+x²+2x+1+y²-10y+25 =13

  2x²+2y²-2x-16y+26 = 0

  x²+y²-x-8y+13 = 0

is the required equation of locus.

Here (x, y) ≠ (2, 3) and (x, y) ≠ (-1, 5).

3. Find the equation of locus of P. If A = (4, 0), B = (-4, 0) and | PA-PB | = 4.

Sol: The given fixed points are A (4, 0) and B (-4, 0).

let P(x, y) be any point on the locus.

The given geometric condition is | PA-PB | = 4..

   PA-PB = ± 4   PA-PB = 4 or PA-PB = - 4

                     let PA-PB = -4

  PA+4 = PB squaring on both sides.

  PB² = PA²+8PA+16

  (x+4)²+y² = (x-4)²+y²+8PA+16

  (x+4)²+y²-(x-4)²-y² = 8PA+16

  x²+8x+16+y²-x²+8x-16-y² = 8PA+16

  16x = 8PA+16

  16x-16 = 8PA  2x-2 = PA squaring on both sides.

  4x²-8x+4 = PA²= x²-8x+16+y²

   3x²-y² = 12

    = 1 is the required equation of locus. It is the equation of Hyperbola.

Similarly we can take        PA-PB = 4     

                                            PA = PB + 4.

we get same equation.

4. If the distance from P to the points (2, 3) and (2, -3) are in the ratio 2 : 3 then find the equation locus of P.

sol: The given two fixed points are A (2, 3) and B (2, -3), let P (x, y) be any point on the locus.

The given geometric condition is

PA : PB = 2 : 3   3 PA = 2PB squaring on both sides

  9PA² = 4PB²  9 [(x-2)² +(y-3)²] = 4[(x-2)² + (y+3)² ]

  9[ x²-4x+4+y²-6y+9 ] = 4[ x²-4x+4+y²+6y+9 ]

  9x²+9y²-36x-54y+117 = 4x²+4y² -16x+24y+52

  5 x² + 5 y² - 20 x - 78 y + 65 = 0 is the required equation of locus.

5. A (5, 3) and B (3, -2) are two fixed points. Find the equation of locus of P, so that the area of triangle PAB is 9.

sol: The two given fixed vertices of the triangle are A (5, 3) and B (3,-2).

         Let P (x, y) be the point on the locus. It is the third vertex of the triangle PAB.

          The given geometrical condition is area of the triangle PAB = 9

              

            The combined equation is

            (5x-2y-1) (5x-2y-37) = 0 is the required equation of the locus.

6. Find the equation of locus of the point which is collinear with the point (3, 4) and (-4, 3).

sol: The given two fixed points are A (3, 4) and B (-4, 3)

          Let P (x, y) be any point on the locus

given geometric condition is the three points A, B, P are collinear.

                 The area of the triangle PAB is zero.

                 

               Is the required equation of the locus.

7. Find the equation of locus of a point which is always equidistant from the points A (a+b, a-b) and B (a-b, a+b)

sol: The given two fixed points are A (a+b, a-b) and B (a-b, a+b). Let P (x, y) be any point on the locus.

Given geometric condition is PA = PB  PA2 = PB2

  (x-(a+b))² + (y-(a-b))² = (x-(a-b))² + (y-(a+b))²

  ((x-a) - b)² + ((y-a)+b)² = ((x-a) + b)² + ((y-a) - b)²

  ((x-a) - b)² - ((x-a) + b)² = ((y-a) - b)² - ((y-a) + b)²

  -4b (x-a) = -4b (y-a)  x-a = y-a  x = y or y = x or x - y = 0

           is the required equation of the locus

8. An iron rod of length 2l is sliding on two mutually perpendicular lines. Find the equation of locus of the mid point of the rod.

9. A straight rod of length 9 slides with its ends, A, B always on the X and Y - axes respectively. Then find the locus of the of centroid of ΔOAB.

10. A (5, 3), B (3, -2) and C (2, -1) are three points. If P is a point such that the area of the quadrilateral PABC is 10, then find the locus of P.

sol: The given three vertices of the quadrilateral are A (5, 3), B (3, -2) and C (2, -1).

         Let P (x, y) be the fourth vertex of the quadrilateral. It is on the locus. Given geometric condition is area of the quadrilateral PABC = 10

Writer P. Balreddy