Question

For an ellipse with axes as coordinate axes, the maximum area of a rectangle that can be inscribed in the ellipse is $16$ sq. units. If $e=\frac{\sqrt{3}}{2}$ then equation of the ellipse is:

• A. $\frac{x^2}{16}+\frac{y^2}{4}=1$
• B. $\frac{x^2}{16}+\frac{y^2}{8}=1$
• C. $\frac{x^2}{64}+\frac{y^2}{32}=1$
• D. $\frac{x^2}{20}+\frac{y^2}{16}=1$

Answer 1

The correct option is A $\frac{x^2}{16}+\frac{y^2}{4}=1$

The vertices of any rectangle inscribed in an ellipse is given by

$(\pm a\cos \theta ,\pm b\sin \theta )$

The area of the rectangle is given by

$A=4ab\sin \theta \cos \theta =2ab\sin \left(2\theta \right)$

Hence, the area is maximum when $\sin \left(2\theta \right)=1$.

The maximum area is $A=2ab$

$\therefore 2ab=16$

$\Rightarrow ab=8$ and $e=\frac{\sqrt{3}}{2}$

And, ${\left(ae\right)}^2=a^2-b^2$

$\Rightarrow b^2=\frac{1}{4}{a}^2$

$\Rightarrow b=\frac{a}{2}$

Using two equations we get,

$a=4$ and $b=2$

$\therefore$ the equation of the ellipse is:

$\frac{x^2}{16}+\frac{y^2}{4}=1$