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

The Correct Option is D

The parametric co-ordinates of point that lies on an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 are \left(a\,cos\,0, b\, sin\, 0\right).$ Let the co-ordinates of the vertices of rectangle $ABCD$ are $A \left(a\, cos \,0, b\, sin\, 0\right), B\left(-a\,cos\,0,\,b\,sin\,0\right) C\left(-a\,cos\,0,\,b\,sin\,0\right)$ and $D\left(a\,cos\,0,\,-b\,sin\,0\right)$ then length of rectangle, $AB = 2a\, cos\, 0$ and breadth of rectangle, $AD =2b\, sin \,0$ \therefore Area of rectangle = AB\times AD $= 2a\, cos\, 0 \times 26 \,sin\, 0$ $\Rightarrow $ Area of rectangle, $A = 2\,ab\, sin \,20\,...\left(i\right)$ $\therefore \frac{dA}{d\theta }=2\times2\,ab\,cos\,2\theta$ On putting $\frac{dA}{d0}=0$ for maxima or minima. $\therefore \frac{dA}{d\theta }=0$ $\Rightarrow cos\,2\theta =0$ $\Rightarrow 2\theta=\frac{\pi}{2} \Rightarrow \theta =\frac{\pi}{4}$ Now $\frac{d^{2}A}{d\theta ^{2}}=-8ab\,sin^{2}\,\theta$ 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 $= 2\,ab\,sq unit. \left(From \left(i\right)\right)$ From E (i) Area of rectangle, $A = 2\,ab\, sin\, 2\theta$ $\because A \,\propto\,sin\,2\theta$ and $-1 \le sin\,2\theta \le1$ $\therefore$ A is maximum when $sin\,2\theta=1$ $\Rightarrow$ Maximum area of rectangle $= 2ab\,sq\, unit.$