Let the maximum area of the triangle that can be inscribed in the...

Plotting the diagram as per given information we get,

Now finding area of the triangle we get,

$A = \dfrac{1}{2}a\left(1 - cos\theta\right)\left(4sin\theta\right)$

$A = 2a\left(1 - cos \theta\right)sin\theta$

Differentiating to get maxima and minima we get,

$\dfrac{dA}{d\theta} = 2a\left(\sin^{2}\theta + cos\theta - \cos^{2}\theta\right)$

$\dfrac{dA}{d\theta} = 0 ⇒ 1 + cos\theta - 2\cos^{2}\theta = 0$

$cos\theta = 1\left(Reject\right)$

OR

$cos\theta = \dfrac{- 1}{2} ⇒ \theta = \dfrac{2\pi}{3}$

$\dfrac{{d}^{2}A}{d{\theta}^{2}} = 2a\left(2\sin^{2}\theta - sin\theta\right)$

$\dfrac{{d}^{2}A}{d{\theta}^{2}} < 0$ for $\theta = \dfrac{2\pi}{3}$

Now, {A}_\max = \dfrac{3\sqrt{3}}{2}a = 6\sqrt{3}

$⇒ a = 4$

Now, $e = \sqrt{\dfrac{{a}^{2} - {b}^{2}}{{a}^{2}}} = \dfrac{\sqrt{3}}{2}$