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Q.
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
Conic Sections
Solution:
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 b e=2 \sqrt{6} $
$\therefore \frac{4}{b^{2}-24}+\frac{16}{b^{2}}=1 $
$\Rightarrow b^{4}-44 b^{2}+16 \times 24=0 $
$\Rightarrow \left(b^{2}-32\right)\left(b^{2}-12\right)=0 $
$\Rightarrow b^{2}=32($ as $ b>4)$
$\Rightarrow a^{2}=8$
