Question

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

Answer 1

Let the equation of the required ellipse be
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ...(i)
The coordinates of its foci are $(\pm ae,0)$,
i.e., $(\pm 3,0)$
$\therefore$ ae=3
$\Rightarrow (ae{)}^2=9$^{2} \) ...(i)^{2} The required ellipse passes through (4,1)
$\therefore \frac{4^2}{a^2}+\frac{1^2}{b^2}=1$
$\Rightarrow 16b^2+a^2=a^2{b}^2$
$\Rightarrow a^2+16b^2=a^2{b}^2$ ...(ii)
Now,
$b^2=a^2(1-e^2)$
$\Rightarrow b^2=a^2-a^2{e}^2$
$\Rightarrow b^2=a^2-9$ [Using equation (i)] ...(iii)
Substituting $b^2=a^2-9$ in equation (ii), we get
$a^2+16(a^2-9)=a^2(a^2-9)$
$\Rightarrow a^2+16a^2-144=a^4-9a^2$
$\Rightarrow 17a^2-144=a^2-9a^2$
$\Rightarrow a^4-9a^2-17a^2+144=0$
$\Rightarrow a^4-26a^2+144=0$
$\Rightarrow a^4)-18a^2-8a^2+144=0$
$\Rightarrow a^2(a^2-18)-8(a^2-8)=0$
$\Rightarrow (a^2-18)(a^2-8)=0$
$\Rightarrow a^2=18$ or, $a^2\ne 8$
$\Rightarrow a^2=18$
Putting $a^2=18$ in equation (iii), we get
$\therefore$ The required equation of the ellipse is $\frac{\frac{x^2}{18}}{\frac{y^2}{9}}=1$