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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