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

Explanation

Let P(h,k) be any point on the locus and let A(0,2) and `B(0,-2)` be the given points.
By the given condition. We get
`PA+PB=6`
or `sqrt((h-0)^2+(k-2)^2)+sqrt((h-0)^2+(k+2)^2)=6`
or `sqrt(h^2+(k-2)^2)=6-sqrt((h-0)^2+(k+2)^2)`
or `h^2+(k-2)^2=36-12sqrt(h^2+(k+2)^2)+h^2+(k+2)^2`
or `-8k-36=-12sqrt(h^2+(k+2)^2)`
`(2k+9)=3sqrt(h^2+(k+2)^2)`
or `(2k+9)^2=9{h^2+(k+2)^2}`
or `4k^2+36k+81=9h^2+9k^2+36k+36`
or `9h^2+5k^2=45`
Hence the locus of (h,k) is `9x^2+5y^2=45`.