Question

Find the locus of a point P such that the sum of its distances from (0,2) and (0,-2) is 6​

Answer 1

Answer:

By distance formula,

[x^2 + (y-2)^2]^1/2 + [x^2 + (y+2)^2]^1/2 = 6.

(x^2 + y^2 - 4y + 4)^1/2 + (x^2 + y^2 + 4y + 4)^1/2 = 6.

Let x^2 + y^2 + 4 = a.

(a - 4y)^1/2 + (a + 4y)^1/2 = 6.

(a - 4y)^1/2 = 6 - (a + 4y)^1/2.

Squaring both sides,

a - 4y = a + 4y + 36 - 12* (a + 4y)^1/2.

8y + 36 = 12* (a + 4y)^1/2.

2y + 9 = 3* (a + 4y)^1/2.

Squaring again,

4y^2 + 36y + 81 = 9a + 36y.

9a - 4y^2 - 81 = 0.

Putting back the original value of a,

9x^2 + 5y^2 - 45 = 0.