Let the maximum area of the triangle that can be inscribed in the ellipse x 2 a 2 + y 2 4 = 1 , a > 2 ,…

Plotting the diagram as per given information we get,

Now finding area of the triangle we get,

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

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

Differentiating to get maxima and minima we get,

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

$\frac{\mathrm{d}A}{ \mathrm{d}\theta }=0\Rightarrow 1+\cos \theta -2{\cos }^2\theta =0$

$\cos \theta =1\left(\mathrm{Reject}\right)$

OR

$\cos \theta =\frac{-1}{2}\Rightarrow \theta =\frac{2\pi }{3}$

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

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

Now, ${\mathrm{A}}_{\max }=\frac{3\sqrt{3}}{2}\mathrm{a}=6\sqrt{3}$

$\Rightarrow a=4$

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