Question

Find the equation of the ellipse whose foci are $(0,\pm 3)$ and length of the minor axis is $16$

Answer 1

Since the foci of the ellipse are $F_1(0,-3)$ and $F_2(0,3)$ which lie on the $y-$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 $y-$axis.

Hence the equation of the ellipse can be taken as

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

Here $c=3$

Also, the length of the minor axis is $16$

$\Rightarrow 2b=16$

$\Rightarrow b=\frac{16}{2}=8$

We know that $b^2=a^2-c^2$

$\Rightarrow 8^2=a^2-3^2$

$\Rightarrow a^2=64+9=73$

From $\left(1\right)$ , the equation of the ellipse is $\frac{x^2}{64}+\frac{y^2}{73}=1$