JEE MAIN - Mathematics (2022 - 24th June Evening Shift - No. 8)
Let the maximum area of the triangle that can be inscribed in the ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 4} = 1,\,a > 2$$, 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 ellipse is :
$${{\sqrt 3 } \over 2}$$
$${1 \over 2}$$
$${1 \over {\sqrt 2 }}$$
$${{\sqrt 3 } \over 4}$$
Explanation
Given ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 4} = 1,\,a > 2$$
_24th_June_Evening_Shift_en_8_1.png)
$$\therefore$$ Let A($$\theta$$) be the area of $$\Delta$$ABB'
Then $$A(\theta ) = {1 \over 2}4\sin \theta (a + a\cos \theta )$$
$$A'(\theta ) = a(2\cos \theta + 2{\cos ^2}\theta )$$
For maxima $$A'(\theta ) = 0$$
$$ \Rightarrow \cos \theta = 1,\,\,\cos \theta = {1 \over 2}$$
But for maximum area $$\cos \theta = {1 \over 2}$$
$$\therefore$$ $$A(\theta ) = 6\sqrt 3 $$
$$ \Rightarrow 2{{\sqrt 3 } \over 2}\left( {a + {a \over 2}} \right) = 6\sqrt 3 $$
$$ \Rightarrow a = 4$$
$$\therefore$$ $$e = \sqrt {1 - {{{b^2}} \over {{a^2}}}} = \sqrt {1 - {4 \over {16}}} = {{\sqrt 3 } \over 2}$$
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