Question

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

• A. $\sqrt{2}$
• B. $\frac{1}{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

Answer: (B)

$\frac{1}{2}$

In an ellipse, the sum of the distance of any point on the ellipse from the focii equals $2a.$

And the distance between the focii equals $2ae.$

$\Rightarrow 2ae=8$ or $ae=4$

Also, $\sqrt{(4\sqrt{2}+4{)}^2+(2\sqrt{6}{)}^2}+\sqrt{(4\sqrt{2}-4{)}^2+(2\sqrt{6}{)}^2}=2a$

$\Rightarrow \sqrt{32+16+24+32\sqrt{2}}+\sqrt{32+16+24-32\sqrt{2}}=2a$

$\Rightarrow 2\sqrt{2}\times \sqrt{9+4\sqrt{2}}+2\sqrt{2}\times \sqrt{9-4\sqrt{2}}=2a$

$\Rightarrow \sqrt{2}(2\sqrt{2}+1\pm (2\sqrt{2}-1\left)\right)=a$

$\Rightarrow a=8$ or $a=2\sqrt{2}$

$\therefore e=\frac{1}{2}$ or $e=\sqrt{2}$

Since the eccentricity of an ellipse is less than $1,e=\frac{1}{2}$