JEE Main 2022 · 24 Jun · Shift 2 · Mathematics Q9

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MathematicsSingle correct+4 / -1Question 9 of 90

Let the maximum area of the triangle that can be inscribed in the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}4=1$ , $a>2$ , 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 $6\sqrt3$ . Then the eccentricity of the ellipse is:

\dfrac{\sqrt3}2

\dfrac12

\dfrac1{\sqrt2}

\dfrac{\sqrt3}4

Answer and worked solution

Official answer: A

Set up. Vertex at $(-a,0)$ ; the vertical side joins $(a\cos\theta,\pm2\sin\theta)$ .

Area. $\Delta=\frac12(4\sin\theta)(a+a\cos\theta)=2a\sin\theta(1+\cos\theta)$ , which is greatest at $\theta=\frac\pi3$ : $\Delta_{\max}=\dfrac{3\sqrt3}2a$ .

Find $a$ . $\dfrac{3\sqrt3}2a=6\sqrt3$ gives $a=4$ .

Eccentricity. $e=\sqrt{1-\dfrac4{16}}=\dfrac{\sqrt3}2$ .

(A)

Topics:Ellipse Maxima and minima

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JEE Main 2022 · 24 Jun · Shift 2, Mathematics question 9