EQUATION OF LOCUS OF A POINT WITH THE GIVEN CONDITION

About "Equation of Locus of a Point With the Given Condition"

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.

Equation of Locus of a Point With the Given Condition - Practice questions

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   

PA =  (x - 3)2 + (y - 5) -----(1)

PB2   =  (x - 1)2 + (y + 1) -----(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)=  (-3)2 + (-8)=  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

After having gone through the stuff given above, we hope that the students would have understood "Equation of Locus of a Point With the Given Condition". 

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