Question

Find the equation of the ellipse whose vertices are $(\pm 6,0)$ and foci are $(\pm 4,0)$

Answer 1

Since the foci of the ellipse are $F_1(-4,0)$ and $F_2(4,0)$ which lie on the $x-$axis and the mid-point of the line segment $F_1{F}_2$ is $(0,0),\therefore ,$ origin is the centre of the ellipse and the major axis lies along $x-$axis.

Hence the equation of the ellipse can be taken as

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ .........$\left(1\right)$

As the vertices are $(\pm 6,0)$.So,$a=6$

Also,$c=4$

We know that $b^2=a^2-c^2=6^2-4^2=36-16=20$

From equation $\left(1\right)$, the equation of the ellipse is $\frac{x^2}{36}+\frac{y^2}{20}=1$