QJEEM22Ma2b2a: JEE Main 2022 Mathematics - Class 11 JEEM22Ma2b2a Solution: Let the maximum area of the triangle that can be inscribed in…

Given ellipse {{{x^2}} \over {{a^2}}} + {{{y^2}} \over 4} = 1,\,a > 2

$\therefore$ 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 $\cos \theta = {1 \over 2}$

$\therefore$ $A(\theta ) = 6\sqrt 3$

$\Rightarrow 2{{\sqrt 3 } \over 2}\left( {a + {a \over 2}} \right) = 6\sqrt 3$

$\Rightarrow a = 4$

$\therefore$ $e = \sqrt {1 - {{{b^2}} \over {{a^2}}}} = \sqrt {1 - {4 \over {16}}} = {{\sqrt 3 } \over 2}$