Question

Let $S_1$ and $S_2$ be the focii of the ellipse $\frac{x^2}{16}+\frac{y^2}{8}=1$. If $A(x,y)$ is any point on the ellipse, then the maximum area of the $△A{S}_1{S}_2$ (in sq units) is

• A. $2\sqrt{2}$
• B. $2\sqrt{3}$
• C. $8$
• D. $4$
• E. $16$

Answer 1

Answer: (C)

$8$

Given equation of ellipse is $\frac{x^2}{16}+\frac{y^2}{8}-1$.
Here, $a^2=16,b^2=8$
Eccentricity of the ellipse is
$e=\sqrt{\frac{a^2-b^2}{a^2}}=\sqrt{\frac{16-8}{16}}$
$\Rightarrow e=\frac{1}{\sqrt{2}},a=4$ and $b=2\sqrt{2}$
Now, area is maximum when vertex $A$ is at $B(0,b)$.
Therefore, maximum area $=\frac{1}{2}\times S{S}^{\prime }\times BC$
$=\frac{1}{2}\times \text{(distance between focii)}\times b$
$=\frac{1}{2}\times 2ae\times b=eab=\frac{1}{\sqrt{2}}\times 4\times 2\sqrt{2}$
$=8$ sq. units