QJEEAdv22Ma0b3f: JEE Advanced 2022 Mathematics - Class 11 JEEAdv22Ma0b3f Solution: Consider the ellipse Let , be a point. A straight line drawn…

Question

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Consider the ellipse

Accepted Answer

Given:

\frac{x^2}{4}+\frac{y^2}{3}=1

Let

\quad \alpha=2 \cos \phi

Tangent at

\mathrm{E}(2 \cos \phi, \sqrt{3} \sin \phi)

to the ellipse is

\frac{x \cos \phi}{2}+\frac{y \sin \phi}{\sqrt{3}}=1

This intersect

x

-axis at

\mathrm{G}(2 \sec \phi, 0)

\begin{aligned}&amp; \text { Area of triangle } \mathrm{FGH}=\frac{1}{2}(2 \sec \phi-2 \cos \phi) 2 \sin \phi \\\\&amp; \Delta=2 \sin ^2 \phi \cdot \tan \phi \\\\&amp; \Delta=(1-\cos 2 \phi) \cdot \tan \phi\end{aligned}

\begin{aligned}\text {I. If }\phi &amp; =\frac{\pi}{4}, \Delta=1 \rightarrow Q \\\\\text {II. If }\phi &amp; =\frac{\pi}{3}, \Delta=2\left(\frac{\sqrt{3}}{2}\right)^2 \cdot \sqrt{3} \\\\&amp; =\frac{3 \sqrt{3}}{2} \rightarrow \mathrm{T}\end{aligned}

\begin{aligned}\text {III. If }\phi &amp; =\frac{\pi}{6}, \Delta=2\left(\frac{1}{2}\right)^2 \cdot \frac{1}{\sqrt{3}} \\\\&amp; =\frac{1}{2 \sqrt{3}} \rightarrow S\end{aligned}

\begin{aligned}\text {IV. If }\phi=\frac{\pi}{12}, \Delta &amp; =\left(1-\frac{\sqrt{3}}{2}\right) \cdot(2-\sqrt{3}) \\\\&amp; =\frac{(\sqrt{3}-1)^4}{8} \rightarrow P\end{aligned}

List-I	List-II
(I) If $\phi=\frac{\pi}{4}$ , then the area of the triangle $F G H$ is	(P) $\frac{(\sqrt{3}-1)^{4}}{8}$
(II) If $\phi=\frac{\pi}{3}$ , then the area of the triangle $F G H$ is	(Q) 1
(III) If $\phi=\frac{\pi}{6}$ , then the area of the triangle $F G H$ is	(R) $\frac{3}{4}$
(IV) If $\phi=\frac{\pi}{12}$ , then the area of the triangle $F G H$ is	(S) $\frac{1}{2 \sqrt{3}}$
	(T) $\frac{3 \sqrt{3}}{2}$

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