Question

Let the maximum area of the triangle that can be inscribed in the ellipse $\frac{x^2}{a^2}+\frac{y^2}{4}=1,a>2$, having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the $y$-axis, be $6\sqrt{3}$. Then the eccentricity of the ellipse is :

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{1}{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{\sqrt{3}}{4}$

Answer 1

Answer: (A)

$A=\frac{1}{2}a(1-\cos \theta )(4\sin \theta )$
$A=2a(1-\cos \theta )\sin \theta$
$\frac{dA}{d\theta }=2a\left({\sin }^2\theta +\cos \theta -{\cos }^2\theta \right)$
$\frac{dA}{d\theta }=0\Rightarrow 1+\cos \theta -2{\cos }^2\theta =0$
$\cos \theta =1$ ( Reject )
OR
$\cos \theta =\frac{-1}{2}\Rightarrow \theta =\frac{2\pi }{3}$
$\frac{d^{2}A}{d{\theta }^2}=2a\left(2{\sin }^2\theta -\sin \theta \right)$
$\frac{d^{2}A}{d{\theta }^2}<0\text{ for }\theta =\frac{2\pi }{3}$
Now, $A_{\max }=\frac{3\sqrt{3}}{2}a=6\sqrt{3}$
$a=4$
Now, $e=\sqrt{\frac{a^2-b^2}{a^2}}=\frac{\sqrt{3}}{2}$