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]
