A point moves so that the sum of its distances from $(a e, 0)$ and $(-a e, 0)$ is $2 \mathrm{a}$, then the equaton to its locus where $b^2=a^2\left(1-e^2\right)$ is

Suppose that the co-ordinates of the moving point be $\left(x,y\right),$then from the given condition

$\sqrt{\left[{\left(x-ae\right)}^2+y^2\right]}+\sqrt{\left[{\left(x+ae\right)}^2+y^2\right]}=2a .....\left(1\right)$

Now

$\left[{\left(x-ae\right)}^2+y^2\right]-\left[{\left(x+ae\right)}^2+y^2\right]=-4aex ....\left(2\right)$

$\left[\because {\left(a-b\right)}^2-{\left(a+b\right)}^2=-4ab\right]$

On dividing $\left(2\right)$by $\left(1\right)$, we get

$\frac{\left[{\left(x-ae\right)}^2+y^2\right]-\left[{\left(x+ae\right)}^2+y^2\right]}{\sqrt{\left[{\left(x-ae\right)}^2+y^2\right]}+\sqrt{\left[{\left(x+ae\right)}^2+y^2\right]}}=\frac{-4aex}{2a}$

$\Rightarrow \frac{{\left[\sqrt{{\left(x-ae\right)}^2+y^2}\right]}^2-{\left[\sqrt{{\left(x+ae\right)}^2+y^2}\right]}^2}{\sqrt{\left[{\left(x-ae\right)}^2+y^2\right]}+\sqrt{\left[{\left(x+ae\right)}^2+y^2\right]}}=-2ex$

$\Rightarrow \sqrt{\left[{\left(x-ae\right)}^2+y^2\right]}-\sqrt{\left[{\left(x+ae\right)}^2+y^2\right]}=-2ex......\left(3\right)$

Adding $\left(1\right)$and $\left(3\right)$ we get

$2\sqrt{\left[{\left(x-ae\right)}^2+y^2\right]}=2\left(a-ex\right).$

Squaring on both sides, we get

$x^2-2aex+a^2{e}^2+y^2=a^2-2aex+e^2{x}^2$

 $\Rightarrow x^2\left(1-e^2\right)+y^2=a^2\left(1-e^2\right)$

$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{a^2\left(1-e^2\right)}=1$

$\therefore \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$