Question

$\text{Find the locus of a point such that the sum of its distances from (0,2) and (0,-2) is 6}.$

Answer 1

$\text{Let p(h,k) be a point. Let the given points be A(0,2) and B(0,-2) According to the given conditin,}AP+BP=6\Rightarrow \sqrt{(h-0{)}^2+(k-2{)}^2}+\sqrt{(h-0{)}^2+(k+2{)}^2}\Rightarrow \sqrt{h^2(k-2)+h^2}=6-\sqrt{h^2+(k+2{)}^2}\text{Squaring both sides,we get:}\Rightarrow h^2+(k-2{)}^2=36+h^2+(k+2{)}^2-12{\sqrt{h^2+(k+2)}}^2\Rightarrow h^2+k^2+4-4k=36+h^2+k^2+4+4k-12\sqrt{h^2+(k+2{)}^2}\Rightarrow 3\sqrt{h^2+(k+2{)}^2}=9+2k\Rightarrow 9(h^2+k^2+4+4k)=81+4k^2+36k\left(Squaringbothsides\right)\Rightarrow 9h^2+9k^2+36+36k=81+4k^2+36k\Rightarrow 9h^2+5k^2-45=0Hence,thelocusof(h,k)is9x^2+5y^2-45=0$