Let the maximum area of the triangle that can be inscribed in the ellipse x^2 a^2+ y^2 4=1,a>2, having one of its vertices at one end of the major…

Plotting the diagram as per given information we get,

Now finding area of the triangle we get,

$$A=\frac{1}{2}a(1-\mathrm{cos}\theta )(4\mathrm{sin}\theta )$$A=21​a(1−cosθ)(4sinθ)

$$A=2a(1-cos\theta )\mathrm{sin}\theta$$A=2a(1−cosθ)sinθ

Differentiating to get maxima and minima we get,

$$\frac{\mathrm{dA}}{d\theta }=2a({\mathrm{sin}}^{2}\theta +\mathrm{cos}\theta -{\mathrm{cos}}^{2}\theta )$$dθdA​=2a(sin2θ+cosθ−cos2θ)

$$\frac{dA}{d\theta }=0\Rightarrow 1+\mathrm{cos}\theta -2{\mathrm{cos}}^{2}\theta =0$$dθdA​=0⇒1+cosθ−2cos2θ=0

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

OR

$$\mathrm{cos}\theta =\frac{-1}{2}\Rightarrow \theta =\frac{2\pi }{3}$$cosθ=2−1​⇒θ=32π​

$$\frac{{d}^{2}A}{d{\theta }^{2}}=2a(2{\mathrm{sin}}^{2}\theta -\mathrm{sin}\theta )$$dθ2d2A​=2a(2sin2θ−sinθ)

$\frac{{d}^{2}A}{d{\theta }^{2}}<0$dθ2d2A​<0 for $\theta =\frac{2\pi }{3}$θ=32π​

Now, ${A}_{\mathrm{max}}=\frac{3\sqrt{3}}{2}a=6\sqrt{3}$Amax​=233​​a=63​

$$\Rightarrow a=4$$⇒a=4

Now, $e=\sqrt{\frac{{a}^{2}-{b}^{2}}{{a}^{2}}}=\frac{\sqrt{3}}{2}$e=a2a2−b2​​=23​​