# EQUATION OF LOCUS OF A POINT EXAMPLES

## About "Equation of Locus of a Point Examples"

Equation of Locus of a Point Examples :

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 Examples - Practice questions

Question 1 :

If O is origin and R is a variable point on y2 = 4x, then find the equation of the locus of the mid-point of the line segment OR.

Solution :

Let the coordinates R be (a, b) and let P (h, k) be the midpoint of OQ. Then

h = (a + 0)/2  =  a/2

k = (0 + b)/2  =  b/2

a  =  2h and b  =  2k

Here a and b are two variables which are to be eliminated. Since (a, b) lies on  y2 = 4x,

b2  =  4a

(2k)2  =  4 (2h)

4k2  =  8h

h = x and k = y

4y2  =  8x

y2  =  2x

Hence the equation of locus y2  =  2x.

Question 2 :

The coordinates of a moving point P are (a/2 (cosec θ + sin θ) , b/2 (cosecθ − sin θ)), where θ is a variable parameter. Show that the equation of the locus P is b2x2 − a2y2 = a2b2 .

Solution :

h  =  a/2 (cosec θ + sin θ)

cosec θ + sin θ  =  2h/a

(cosec θ + sin θ)2  =  (2h/a)2

cosecθ + sin2θ + 2cosecθ sin θ  =  4h2/a2

cosecθ + sin2θ + 2(1/sin θ) sin θ  =  4h2/a2

cosecθ + sin2θ + 2  =  4h2/a------(1)

k  =  b/2 (cosec θ - sin θ)

cosec θ - sin θ  =  2k/b

(cosec θ - sin θ)2  =  (2k/b)2

cosecθ + sin2θ - 2cosecθ sin θ  =  4k2/b2

cosecθ + sin2θ - 2(1/sin θ) sin θ  =  4k2/b2

cosecθ + sin2θ - 2  =  4k2/b2  ------(2)

(1) - (2)

(cosec2θ+sin2θ+2)-(cosec2θ+sin2θ-2) = (4h2/a2) - (4k2/b2)

4  =  (4h2/a2) - (4k2/b2)

Divide by 4 on both sides.

1  =  (h2/a2) - (k2/b2)

a2b2  =  h2b2 - k2a2

Replacing h and k by x and y respectively.

a2b2  =  x2b2 - y2a2

Question 3 :

If P(2,−7) is a given point and Q is a point on 2x2 + 9y2 = 18, then find the equations of the locus of the mid-point of PQ.

Solution :

Let Q be (a, b)

Midpoint of the line segment PQ :

Let A be the midpoint of the line segment PQ (h, k)

P (2, - 7) and Q (a, b)

Midpoint of PQ  =  (x1 + x2) /2, (x1 + x2) /2

(h , k)  =  (2 + a)/2, (-7 + b)/2

 (2 + a) /2  =  h2 + a   =  2 ha = 2h - 2 (-7 + b) /2  =  k-7 + b   =  2 kb = 2k + 7

Since the point Q lies on the curve

2x2 + 9y2 = 18

2a2 + 9b2 = 18  -----(1)

By applying the values of a and b in the above equation, we get

2(2h - 2)2 + 9(2k + 7)2  =  18

2(4h2 - 8h + 4) + 9(4k2 + 28k + 49)  =  18

8h2 - 16h + 8 + 36k2 + 72k + 441  =  18

8h2 + 36k2- 16h + 72k + 449 - 18  =  0

8h2 + 36k2- 16h + 72k + 431  =  0

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

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