Question

If the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is inscribed in a rectangle whose length to breadth ratio is $2:1$ in such a manner that it touches the sides of rectangle, then the area of the rectangle is

• A. $4\left(\frac{a^2+b^2}{7}\right)$
• B. $4\left(\frac{a^2+b^2}{3}\right)$
• C. $12\left(\frac{a^2+b^2}{5}\right)$
• D. $8\left(\frac{a^2+b^2}{5}\right)$

Answer 1

The correct option is D $8\left(\frac{a^2+b^2}{5}\right)$

As given that, ellipse is inscribed in a rectangle so the length of major axis is equal to the length of rectangle and length of minor axis is equal to the breadth of rectangle.
$\therefore \frac{2a}{2b}=\frac{2}{1}\Rightarrow a=2b\cdots \left(1\right)$

Area of rectangle
$=2a\times 2b=4ab=8b^2\left(\text{from equation}\left(1\right)\right)$
Now, check
$\Rightarrow 8\left(\frac{4b^2+b^2}{5}\right)=8b^2$