Question

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

Answer 1

Let point be P(h,k)
((h – 0)^2 + (k – 2)^2)^1/2 + ((h – 0)^2 + (k + 2)^2)^1/2 = 6
(h^2 + (k – 2)^2)^1/2 + (h^2 + (k + 2)^2)^1/2 = 6
(h^2 + (k – 2)^2)^1/2 = 6 – (h^2 + (k + 2)^2)^1/2
now square both sides
h^2 + k^2 – 4k + 4 = 36 + h^2 + k^2 + 4k + 4 – 12(h^2 + (k + 2)^2)^1/2
– 4k = 36 + 4k – 12(h^2 + (k + 2)^2)^1/2
8k + 36 = 12(h^2 + (k + 2)^2)^1/2
2k + 9 = 3((h^2 + (k + 2)^2)^1/2)
square both sides
4k^2 + 81 + 36k = 9(h^2 + k^2 + 4 + 4k)
9h^2 + 5k^2 = 45
9x^2 + 5y^2 = 45