Answer
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{a^2(1-e^2)}=1$
Work Step by Step
We need to write $b$ in terms of $c$.
$b^2=a^2-c^2$
$c=ae$,
Therefore, $b^2=a^2-(ae)^2=a^-a^2e^2$
or, $b^2=a^2(1-e^2)$
The standard form of the equation of an ellipse is:
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
or, $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{a^2(1-e^2)}=1$
