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

The correct option is C $9x^2+5y^2-28y-45=0$
Let the point P be (x,y)

According to question,

$\sqrt{x^2+(y-2{)}^2}+\sqrt{x^2+(y+2{)}^2}=6$

$\Rightarrow \sqrt{x^2+(y-2{)}^2}-6=\sqrt{x^2+(y+2{)}^2}$

On squaring both sides,

$\Rightarrow x^2+(y-2{)}^2+36-12\sqrt{x^2+(y-2{)}^2}=x^2+(y+2{)}^2$

$\Rightarrow 36+y^2+4-4y=12\sqrt{x^2+(y-2{)}^2}+y^2+4+4y$

$\Rightarrow 36-4y=12\sqrt{x^2+(y-2{)}^2}+4y$

$\Rightarrow 36-8y=12\sqrt{x^2+(y-2{)}^2}$

$\Rightarrow 9-2y=3\sqrt{x^2+(y-2{)}^2}$

Again on squaring both sides,

$\Rightarrow 81+4y^2-8y=9x^2+9y^2+36-36y$

$\Rightarrow 9x^2+5y^2-28y-45=0$, which is the required locus.