Question

The equation of ellipse circumscribing the quadrilateral whose sides are given by $x=\pm 2$ and $y=\pm 4$ and distance between foci is $4\sqrt{6}$, is

• A. $\frac{x^2}{32}+\frac{y^2}{8}=1$
• B. $\frac{x^2}{8}+\frac{y^2}{32}=1$
• C. $\frac{x^2}{16}+\frac{y^2}{8}=1$
• D. $\frac{x^2}{8}+\frac{y^2}{16}=1$

Answer 1

Answer: (B)

Let equation be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where $a^2=b^2\left(1-e^2\right)$

Ellipse passes through the point $(2,4)$.
$\therefore \frac{4}{a^2}+\frac{16}{b^2}=1$
Also distance between foci is $4\sqrt{6}$.
$\therefore be=2\sqrt{6}$
$\therefore \frac{4}{b^2-24}+\frac{16}{b^2}=1$
$\Rightarrow b^4-44b^2+16\times 24=0$
$\Rightarrow \left(b^2-32\right)\left(b^2-12\right)=0$
$\Rightarrow b^2=32($ as $b>4)$
$\Rightarrow a^2=8$