An ellipse passing through $(4 \sqrt{2}, 2 \sqrt{6})$ has foci at $(-4,0)$ and $(4,0)$. Then, its eccentricity is

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Aicharya · Accepted Answer

The $y$-coordinate of foci is zero $	herefore$ Major axis is on $\mathrm{X}$-axis $ 	herefore \quad 	ext { ae }=4 $ Let, equation of ellipse be $ \begin{aligned} & \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \ & 	herefore \quad \frac{(4 \sqrt{2})^2}{a^2}+\frac{(2 \sqrt{6})^2}{a^2-16} \ & {\left[\because b^2=a^2\left(1-e^2ight)=a^2-16ight.} \ & \Rightarrow \quad \frac{32}{a^2}+\frac{24}{a^2-16}=1 \ & \Rightarrow \quad 32 a^2-512+24 a^2=a^2\left(a^2-16ight) \ & \Rightarrow \quad 56 a^2-512=a^4-16 a^2 \ & \Rightarrow \quad a^4-72 a^2+512=0 \ & \Rightarrow \quad a^2-64 a^2-8 a^2+512=0 \ & \Rightarrow \quad a^2\left(a^2-64ight)-8\left(a^2-64ight)=0 \ & \Rightarrow \quad\left(a^2-8ight)\left(a^2-64ight)=0 \ & \Rightarrow \quad a^2=64 \Rightarrow a=8 \ & \left(\because a^2=8ight. 	ext { is not possible) } \ & \because \quad 	ext { ae }=4 \Rightarrow 8 	imes e=4 \ & \end{aligned} $ $\Rightarrow \quad e=\frac{1}{2}$

An ellipse passing through $(4 \sqrt{2}, 2 \sqrt{6})$ has foci at $(-4,0)$ and $(4,0)$. Then, its…

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