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

The correct option is C $9x^2+5y^2=45$

The locus of points, whose sum of distance from two fixed points is constant is an ellipse.

For ellipse, $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

Here, the constant sum $2a=6$ or $a=3$

And distance between focii $(0,2)$ and $(0,-2)$ is $4.$

$\Rightarrow 2ae=4$ or $e=\frac{2}{3}$

Since the focii lie on the $y$ axis, the equation of the ellipse will be of the form $\frac{x^2}{a^2(1-e^2)}+\frac{y^2}{a^2}=1$

$\Rightarrow \frac{x^2}{9(1-\frac{4}{9})}+\frac{y^2}{9}=1$

$\Rightarrow \frac{x^2}{5}+\frac{y^2}{9}=1$

i.e. $9x^2+5y^2=45$