Question

Find the equation of the ellipse whose foci are at the points $(2,4)$ and $(2,8)$ and semiminor axis is $3$.

Answer 1

Given the foci of the ellipse are $(2,4)$ and $(2,8)$.

Then centre of the ellipse will be $(\frac{2+2}{2},\frac{4+8}{2})=(2,6)$.

Then equation of the ellipse is $\frac{(x-2{)}^2}{a^2}+\frac{(y-6{)}^2}{b^2}=1$.......(1).

According to the problem,

$b=3$.

Now distance between the foci $=2ae=\sqrt{(2-2{)}^2+(4-8{)}^2}=4$.

Then $ae=2$.....(2).

Now we've $b^2=a^2-(ae{)}^2$ [ Using (2)]

or, $a^2=b^2+(ae{)}^2$

or, $a^2=3^2+2^2$

or, $a^2=13$.

Then equation of the ellipse is $\frac{(x-2{)}^2}{13}+\frac{(y-6{)}^2}{9}=1$.