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 $6\sqrt3$. Then the eccentricity of the ellipse is

• A. $\frac{\sqrt3}{2}$
• B. $\frac{1}{2}$
• C. $\frac{1}{\sqrt2}$
• D. $\frac{\sqrt3}{4}$

Answer 1

Answer: (A)

$\frac{\sqrt3}{2}$

Answer 2

$\frac{x^2}{a^2}+\frac{y^2}{4} = 1, a>2$

Assuming $A(θ)$ be the area of $ΔABB′$

$A(\theta) = \frac{1}{2} 4 sin \theta (a+ a cos\theta)$ and $A' (\theta) = a(2cos\theta+2cos2\theta)$

$⇒ cos\theta = -1, cos\theta = \frac{1}{2}$

$cos\theta = \frac{1}{2}$

$⇒ 2 \frac{\sqrt3}{2} \bigg(a+\frac{a}{2}\bigg) = 6√3$

$⇒ a = 4$

$∴ e = \sqrt{ 1- \frac{b^2}{a^2}}$

$=\sqrt{1-\frac{4}{16}}$

$= \frac{\sqrt3}{2}$

Hence, the correct option is (A): $\frac{\sqrt3}{2}$