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)
PA^{2} + PB^{2} = 20
PA^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2 }
PA^{2 } = (x - 3)^{2} + (y - 5)^{2 } -----(1)
PB^{2} = (x - 1)^{2} + (y + 1)^{2 } -----(2)
(1) + (2) = 20
x^{2} - 6x + 9 + y^{2} - 10y + 25 + x^{2} - 2x + 1 + y^{2} + 2y + 1 = 20
2x^{2} - 8x + 2y^{2} - 8y + 36 = 20
2x^{2} - 8x + 2y^{2} - 8y + 16 = 0
Divide the equation by 2, we get
= x^{2} - 4x + y^{2} - 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
AB^{2} = PA^{2} + PB^{2}
P (x, y) A (1,−6)
PA^{2} = (x - 1)^{2} + (y + 6)^{2}
= x^{2} - 2x + 1 + y^{2} + 12y + 36 --(1)
P (x, y) B (4,−2)
PB^{2} = (x - 4)^{2} + (y + 2)^{2}
= x^{2} - 8x + 16 + y^{2} + 4y + 4 --(2)
A(1,−6) and B(4,−2)
AB^{2} = (1 - 4)^{2} + (-6 - 2)^{2 }= (-3)^{2} + (-8)^{2 }= 9 + 64 = 73
AB^{2} = PA^{2} + PB^{2}
73 = 2x^{2} + 2y^{2} -10x + 16y + 57
2x^{2} + 2y^{2} -10x + 16y + 57 - 73 = 0
2x^{2} + 2y^{2} -10x + 16y - 16 = 0
x^{2} + y^{2} - 5x + 8y - 8 = 0
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