Question

Find the equation for the ellipse that satisfies the given condition:

Vertices $(\pm 5,0)$, foci $(\pm 4,0).$

Answer 1

It is given that vertices $(\pm 5,0)$ foci $(\pm 4,0)$
Clearly, the vertices are on the $x$-axis.
Therefore, the equation of the ellipse will be of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
where $a$ is the semi-major axis and $a=5,ae=4$

$\Rightarrow e=\frac{4}{5}$, where $e$ is eccentricity of the ellipse.
Also, we know that $b^2=a^2(1-e^2)=a^2-a^2{e}^2$
$\therefore b^2=5^2-4^2=9=3^2$

$\Rightarrow b=3$
Thus, the equation of the ellipse is $\frac{x^2}{5^2}+\frac{y^2}{3^2}=1$ or $\frac{x^2}{25}+\frac{y^2}{9}=1$