Question

Area of the greatest rectangle that can be inscribed in the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is :

• A. $\frac{a}{b}$
• B. $\sqrt{ab}$
• C. $ab$
• D. $2ab$

Answer 1

Answer: (D)

Key Idea : The parametric co-ordinates of point that lies on an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1are\left(acos0,bsin0\right).$
Let the co-ordinates of the vertices of rectangle
$ABCD$ are $A\left(acos0,bsin0\right),B\left(-acos0,bsin0\right)C\left(-acos0,bsin0\right)$ and $D\left(acos0,-bsin0\right)$ then
length of rectangle, $AB=2acos0$
and breadth of rectangle, $AD=2bsin0$
\therefore Area of rectangle = AB\times AD
$=2acos0\times 26sin0$
$\Rightarrow$ Area of rectangle, $A=2absin20...\left(i\right)$
$\therefore \frac{dA}{d\theta }=2\times 2abcos2\theta$
On putting $\frac{dA}{d0}=0$ for maxima or minima.
$\therefore \frac{dA}{d\theta }=0$
$\Rightarrow cos2\theta =0$
$\Rightarrow 2\theta =\frac{\pi }{2}\Rightarrow \theta =\frac{\pi }{4}$
Now $\frac{d^{2}A}{d{\theta }^2}=-8absi{n}^2\theta$
Now ${\left(\frac{d^{2}A}{d{\theta }^2}\right)}_{{}_{\theta =\frac{\pi }{4}}}<0$
$\therefore$ Area is maximum at $\theta =\frac{\pi }{4}$
$\Rightarrow$ Maximum Area of rectangle $=2absqunit.\left(From\left(i\right)\right)$
Alternate Solution
From Eq. (i)
Area of rectangle, $A=2absin2\theta$
$\because A\propto sin2\theta$ and $-1\le sin2\theta \le 1$
$\therefore$ A is maximum when $sin2\theta =1$
$\Rightarrow$ Maximum area of rectangle $=2absqunit.$