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Let the maximum area of the triangle that can be inscribed in the ellipse $\mathrm{\frac{x^{2}}{a^{2}}+\frac{y^{2}}{4}=1,a>2}$ , 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 $\mathrm{6\sqrt{3}}$ . Then the eccentricity of the ellipse is:
Option: 1
$\mathrm{\frac{\sqrt{3}}{2}}$

Option: 2
$\mathrm{\frac{1}{2}}$

Option: 3
$\mathrm{\frac{1}{\sqrt{2}}}$

Option: 4
$\mathrm{\frac{\sqrt{3}}{4}}$

Answers (1)

best_answer

Area of $\mathrm{\triangle \underset{(A)}{A B C}=\frac{1}{2} \times 4 \sin \theta \times a(1-\cos \theta)}$

for maximum area,

$\mathrm {\frac{d A}{d \theta}=0}$

$\mathrm {2 a \cos \theta(1-\cos \theta)+2 a \sin \theta(\sin \theta)=0}\\ \mathrm{2a \left(\sin ^{2} \theta-\cos ^{2} \theta+\cos \theta\right)=0}\\ \mathrm{2 a\left(\sin ^{2} \theta-\cos ^{2} \theta+\cos \theta\right)=0}\\ \mathrm{-2 \cos ^{2} \theta+\cos \theta+1=0}$

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

$\mathrm {\theta=0 \quad \theta=\frac{2 \pi}{3}}$

(not possibla)

$\mathrm {\frac{d^{2} A}{d \theta^{2}}=2 a(4 \sin \theta \cos \theta-\sin \theta)}$

at $\mathrm {\theta=\frac{2 \pi}{3} ; \quad \frac{d^{2} A}{d \theta^{2}}= negative }$

Hence at $\mathrm {\theta=\frac{2 \pi}{3}}$ , Area is maximum

maximum Area = $\mathrm {=\frac{1}{2} \times 4 \times \frac{\sqrt{3}}{2} \times a\left(1+\frac{1}{2}\right)=6 \sqrt{3}\\}$

$\mathrm {a \sqrt{3}\left(\frac{3}{2}\right)=6 \sqrt{3}}$

$\mathrm {a=4}$

$\mathrm {\therefore b^{2}=a^{2}\left(1-e^{2}\right)}$

$\mathrm {4=16\left(1-e^{2}\right)}\\$

$\mathrm {e=\frac{\sqrt{3}}{2}}\\$

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