Question

Find the equation of the ellipse referred to its centre whose foci are the points $(4,0)$ and $(4,0)$ and whose eccentricity is $\frac{1}{3}$.

Answer 1

Then center will be (0,0).

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Now foci are $(ae,0),(-ae,0)$

So, $ae=4$

And given

$e=\frac{1}{3}$

So

$a=12$

Also

$e=\sqrt{1-\frac{b^2}{a^2}}$

$\Rightarrow \frac{1}{9}=1-\frac{b^2}{144}$

$\Rightarrow b^2=128$

We have the equation,

$\frac{x^2}{144}+\frac{y^2}{128}=1$

i.e. $8x^2+9y^2=1152$