Question

Find the equation of the ellipse whose vertices are $(\pm 5,0)$ and foci are $(\pm 3,0)$

Answer 1

Since the foci of the ellipse are $F_1(-3,0)$ and $F_2(3,0)$ 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 $(\pm 5,0)$.So,$a=5$

Also,$c=3$

We know that $b^2=a^2-c^2=5^2-3^2=25-9=16$

From equation $\left(1\right)$, the equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$