Question

Find the equation of the ellipse whose axes are along the coordinate axes, vertices are $(\pm 5,0)$ and foci at $(\pm 4,0)$.

Answer 1

Let the equation of the required ellipse be

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

The coordinates of its vertices and foci are $(\pm a,0)$ and $(\pm ae,0)$ respectively.

But, the coordinates of vertices and foci are given as $(\pm 5,0)$ and $(\pm 4,0)$.

Therefore, $a=5$ and $ae=4$

$e=\frac{4}{5}$

Now, $b^2=a^2(1-e^2)$

$b^2=25(1-\frac{16}{25})=9$

Substituting the values of $a^2$ and $b^2$, we obtain $\frac{x^2}{25}+\frac{y^2}{9}=1$, which is the equation of the required ellipse.