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

cosec² θ + sin²θ + 2cosecθ sin θ  =  4h²/a²

cosec² θ + sin²θ + 2(1/sin θ) sin θ  =  4h²/a²

cosec² θ + sin²θ + 2  =  4h²/a²  ------(1) 

k  =  b/2 (cosec θ - sin θ)

cosec θ - sin θ  =  2k/b

(cosec θ - sin θ)²  =  (2k/b)²

cosec² θ + sin²θ - 2cosecθ sin θ  =  4k²/b²

cosec² θ + sin²θ - 2(1/sin θ) sin θ  =  4k²/b²

cosec² θ + 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 

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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