Question

Find the locus of the point $P$ from distance $A(0,2)$ and $B(0,-2)$ is $6$.
$PA+PB=6$

Answer 1

Let the co-ordinate of $P$ be $(h,k)$.

Given

$PA+PB=6$

or, $\sqrt{h^2+(k-2{)}^2}+\sqrt{h^2+(k+2{)}^2}=6$

or, $\sqrt{h^2+(k+2{)}^2}=6-\sqrt{h^2+(k-2{)}^2}$

Squaring both sides we get,

or, $h^2+(k+2{)}^2=36-12\sqrt{h^2+(k-2{)}^2}+h^2+(k-2{)}^2$

or, $(k+2{)}^2-(k-2{)}^2-36=-12\sqrt{h^2+(k-2{)}^2}$

or, $8k-36=-12\sqrt{h^2+(k-2{)}^2}$

or,$(2k-9)=-3\sqrt{h^2+(k-2{)}^2}$

or, $(2k-9{)}^2=9\left\{\right(h^2+(k-2{)}^2\}$

or, $4k^2-36k+81=9h^2+9k^2-36k+36$

or, $5k^2+9h^2=45$

or, $\frac{h^2}{5}+\frac{k^2}{9}=1$.

So, locus of $P$ is $\frac{x^2}{5}+\frac{y^2}{9}=1$.