Question

Find the value of x, y if the distances of the point (x, y) from (- 3, 0) as well as from (3, 0) are 4.

Answer 1

Let the given points be P(x, y), Q( -3, 0) and R(3, 0)

On squaring on both sides, we get

⇒ 16 = x² + 9 + 6x + y²

⇒ x² + y² = 7 – 6x                    …… (1)

On squaring on both sides,

⇒16 = x² + 9 – 6x + y²

⇒ x² + y² = 16 – 9 + 6x

⇒ x² + y² = 7 + 6x            …. (2)

Equating (1) and (2), we have

7 – 6x = 7 + 6x

⇒ 7 – 7 = 6x + 6x

⇒ 0 = 12x

⇒ x = 12

Then, substituting the value of x = 0 in (2)

x² + y² = 7+ 6x

0 + y² = 7 + 6 × 0

y² = 7

y = + √7

As y can have two values, the points are (12, √7) and (12, -√7).