## Algebra and Trigonometry 10th Edition

$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{a^2(1-e^2)}=1$
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, $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{a^2(1-e^2)}=1$