Question

Find the equation of an ellipse whose foci are at $(\pm 3,0)$ and which passes through $(4,1)$.

Answer 1

Let the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

Then, $ae=3$ and $\frac{16}{a^2}+\frac{1}{b^2}=1$

$\Rightarrow \frac{16}{a^2}+\frac{1}{a^2-a^2{e}^2}=1\Rightarrow \frac{16}{a^2}+\frac{1}{a^2-9}=1$

$\Rightarrow a^4-26a^2+144=0\Rightarrow (a^2-18)(a^2-8)=0\Rightarrow a^2=18,a^2=8$.

$\therefore b^2=a^2(1-e{)}^2$ and $ae=3\Rightarrow b^2=a^2-9$.

Now, $a^2=18\Rightarrow b^2=18-9=9$

$a^2=8\Rightarrow b^2=8-9=-1$, which is not possible.

Hence, the equation of the ellipse is $\frac{x^2}{18}+\frac{y^2}{9}=1$.