Question

The area of greatest rectangle that can be inscribed in an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, is

• A. $ab$ sq units
• B. $\frac{ab}{2}$ sq units
• C. $2ab$ sq units
• D. $3ab$ sq units

Answer 1

Answer: (C)

Let $ABCD$ be the rectangle of maximum area with sides $AB=2x$ and $BC=2y$, where $C(x,y)$ is a point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ as shown in the figure. The area $A$ of the rectangle is $4xy$ i.e., $A=4xy$ which gives $A^2=16x^2{y}^2=s$ (say)
Therefore, $s=16x^2\left(1-\frac{x^2}{a^2}\right)\cdot b^2=\frac{16b^2}{a^2}\left(a^2{x}^2-x^4\right)$

$\Rightarrow \frac{ds}{dx}=\frac{16b^2}{a^2}\cdot \left[2a^{2}x-4x^3\right]$
Again, $\frac{ds}{dx}=0\Rightarrow x=\frac{a}{\sqrt{2}}$ and $y=\frac{b}{\sqrt{2}}$
Now, $\frac{d^{2}s}{d{x}^2}=\frac{16b^2}{a^2}\left[2a^2-12x^2\right]$
At $x=\frac{a}{\sqrt{2}},\frac{d^{2}s}{d{x}^2}=\frac{16b^2}{a^2}\left[2a^2-6a^2\right]$
$=\frac{16b^2}{a^2}\left(-4a^2\right)<0$
Thus at $x=\frac{a}{\sqrt{2}},y=\frac{b}{\sqrt{2}},s$ is maximum and hence the area $A$ is maximum.
Maximum area $=4\cdot x\cdot y\cdot =4\cdot \frac{a}{\sqrt{2}}\cdot \frac{b}{\sqrt{2}}=2ab$ sq units.