Question

Find the equation of the ellipse whose foci are $(0,\pm 5)$ and length of the minor axis is $24$

Answer 1

Since the foci of the ellipse are $F_1(0,-5)$ and $F_2(0,5)$ 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 $c=5$

Also, the length of the minor axis is $24$

$\Rightarrow 2b=24$

$\Rightarrow b=\frac{24}{2}=12$

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

$\Rightarrow {12}^2=a^2-5^2$

$\Rightarrow a^2=144+25=169$

From $\left(1\right)$ , the equation of the ellipse is $\frac{x^2}{144}+\frac{y^2}{169}=1$