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)

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