Question

Consider an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ What is the area included between the ellipse and the greatest rectangle inscribed in the ellipse?

• A. $ab(\pi -1)$
• B. $2ab(\pi -1)$
• C. $ab(\pi -2)$
• D. None of the above

Answer 1

The correct option is C $ab(\pi -2)$

As the rectangle lies inside a standard ellipse, the vertices of the rectangle can be assumed to be $(h,k),(h,-k),(-h,k)$ and $(-h,-k)$ without any loss of generality.

$\therefore \frac{h^2}{a^2}+\frac{k^2}{b^2}=1⟹k=b\sqrt{1-\frac{h^2}{a^2}}$

Clearly, the sides of the rectangle based on the coordinates chosen can be determined as $2h$ and $2k$.

$\therefore$ area of rectangle $=A\left(h\right)=2h\times 2k=4hb\sqrt{1-\frac{h^2}{a^2}}⟹4b\sqrt{h^2-\frac{h^4}{a^2}}$

Now, let $P\left(h\right)=h^2-\frac{h^4}{a^2}$. Thus, the maximization of $A\left(h\right)$ boils down just to the maximization of $P\left(h\right)$.

Differentiating $P\left(h\right)$ wrt $h$, we have

$2h-\frac{4h^3}{a^2}=0⟹h=\frac{2h^3}{a^2}⟹2h^2=a^2⟹h=\frac{a}{\sqrt{2}}⟹k=\frac{b}{\sqrt{2}}$

Hence, $A_{max}=2ab$

$\therefore$ area included between this ellipse and rectangle $=\pi ab-2ab=ab(\pi -2)$.

this is the required answer.