Equation of Locus of a Point With the Given Condition :
Here we are going to see how to find equation of locus of a Point with the given condition.
Question 1 :
Find the equation of the locus of a point such that the sum of the squares of the distance from the points (3, 5), (1,−1) is equal to 20
Solution :
Let P (x, y) be the moving point
Let A (3, 5) and B (1,−1)
PA2 + PB2 = 20
PA2 = (x2 - x1)2 + (y2 - y1)2
PA2 = (x - 3)2 + (y - 5)2 -----(1)
PB2 = (x - 1)2 + (y + 1)2 -----(2)
(1) + (2) = 20
x2 - 6x + 9 + y2 - 10y + 25 + x2 - 2x + 1 + y2 + 2y + 1 = 20
2x2 - 8x + 2y2 - 8y + 36 = 20
2x2 - 8x + 2y2 - 8y + 16 = 0
Divide the equation by 2, we get
= x2 - 4x + y2 - 4y + 8
Question 2 :
Find the equation of the locus of the point P such that the line segment AB, joining the points A(1,−6) and B(4,−2), subtends a right angle at P.
Solution :
Let P (x, y) be the moving point
AB2 = PA2 + PB2
P (x, y) A (1,−6)
PA2 = (x - 1)2 + (y + 6)2
= x2 - 2x + 1 + y2 + 12y + 36 --(1)
P (x, y) B (4,−2)
PB2 = (x - 4)2 + (y + 2)2
= x2 - 8x + 16 + y2 + 4y + 4 --(2)
A(1,−6) and B(4,−2)
AB2 = (1 - 4)2 + (-6 - 2)2 = (-3)2 + (-8)2 = 9 + 64 = 73
AB2 = PA2 + PB2
73 = 2x2 + 2y2 -10x + 16y + 57
2x2 + 2y2 -10x + 16y + 57 - 73 = 0
2x2 + 2y2 -10x + 16y - 16 = 0
x2 + y2 - 5x + 8y - 8 = 0
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