Let the maximum area of the triangle that can | vedclass

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Question

$A=\frac{1}{2} a(1-\cos 	heta)(4 \sin 	heta)$

$A =2 a (1-\cos 	heta) \sin 	heta$

$\frac{ dA }{ d 	heta}=2 a \left(\sin ^{2} 	heta+\cos 	h

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2}$

Vedclass · Answer

$A=\frac{1}{2} a(1-\cos 	heta)(4 \sin 	heta)$

$A =2 a (1-\cos 	heta) \sin 	heta$

$\frac{ dA }{ d 	heta}=2 a \left(\sin ^{2} 	heta+\cos 	heta-\cos ^{2} 	hetaight)$

$\frac{ dA }{ d 	heta}=0 \Rightarrow 1+\cos 	heta-2 \cos ^{2} 	heta=0$

$\cos 	heta=1 	ext { (Reject })$

OR

$\cos 	heta=\frac{-1}{2} \Rightarrow 	heta=\frac{2 \pi}{3}$

$\frac{ d ^{2} A }{ d 	heta^{2}}=2 a \left(2 \sin ^{2} 	heta-\sin 	hetaight)$

$\frac{ d ^{2} A }{ d 	heta^{2}}<0 	ext { for } 	heta=\frac{2 \pi}{3}$

Now, $A _{\max }=\frac{3 \sqrt{3}}{2} a =6 \sqrt{3}$

$a=4$

Now, $e =\sqrt{\frac{ a ^{2}- b ^{2}}{ a ^{2}}}=\frac{\sqrt{3}}{2}$

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 ellispe is

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