Question

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

Answer 1

Since the vertices of the given ellipse are on the x-axis, so it is a horizontal ellipse.

Let the equation of the ellipse be

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ where $a^2>b^2.$

Its vertices are $(\pm a,0)$ and therefore, $a=5.$

Its foci are $(\pm c,0)$ and therefore, $c=4.$

$\therefore c^2=(a^2-b^2)\Rightarrow b^2=(a^2-c^2)=(25-16)=9.$

Thus, $a^2=5^2=25$ and $b^2=9.$

Hence, the required equation is $\frac{x^2}{25}+\frac{y^2}{9}=1.$