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
