Consider the ellipse frac{x^2}{4}+frac{y^2}{3 | vedclass

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(C) Let $F(2\cos\phi, 2\sin\phi)$ and $E(2\cos\phi, \sqrt{3}\sin\phi)$.
The equation of the tangent at $E$ is $\frac{x(2\cos\phi)}{4} + \frac{y(\sqrt

Vedclass · Accepted Answer

$(I) \rightarrow (Q); (II) \rightarrow (T); (III) \rightarrow (S); (IV) \rightarrow (P)$

Vedclass · Answer

(C) Let $F(2\cos\phi, 2\sin\phi)$ and $E(2\cos\phi, \sqrt{3}\sin\phi)$.
The equation of the tangent at $E$ is $\frac{x(2\cos\phi)}{4} + \frac{y(\sqrt{3}\sin\phi)}{3} = 1$,which simplifies to $\frac{x\cos\phi}{2} + \frac{y\sin\phi}{\sqrt{3}} = 1$.
Setting $y=0$,we find the $x$-coordinate of $G$ as $x_G = \frac{2}{\cos\phi}$. Thus,$G = (\frac{2}{\cos\phi}, 0)$.
The coordinates are $F(2\cos\phi, 2\sin\phi)$,$G(\frac{2}{\cos\phi}, 0)$,and $H(2\cos\phi, 0)$.
The area of $\Delta FGH = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes HG 	imes FH$.
$HG = |\frac{2}{\cos\phi} - 2\cos\phi| = 2\frac{1-\cos^2\phi}{\cos\phi} = 2\frac{\sin^2\phi}{\cos\phi}$.
$FH = 2\sin\phi$.
Area $A(\phi) = \frac{1}{2} 	imes (2\frac{\sin^2\phi}{\cos\phi}) 	imes (2\sin\phi) = 2	an\phi \sin^2\phi$.
$(I)$ For $\phi = \frac{\pi}{4}$,$A = 2	an(\frac{\pi}{4}) \sin^2(\frac{\pi}{4}) = 2(1)(\frac{1}{2}) = 1 ightarrow (Q)$.
$(II)$ For $\phi = \frac{\pi}{3}$,$A = 2	an(\frac{\pi}{3}) \sin^2(\frac{\pi}{3}) = 2(\sqrt{3})(\frac{3}{4}) = \frac{3\sqrt{3}}{2} ightarrow (T)$.
$(III)$ For $\phi = \frac{\pi}{6}$,$A = 2	an(\frac{\pi}{6}) \sin^2(\frac{\pi}{6}) = 2(\frac{1}{\sqrt{3}})(\frac{1}{4}) = \frac{1}{2\sqrt{3}} ightarrow (S)$.
$(IV)$ For $\phi = \frac{\pi}{12}$,$A = 2	an(\frac{\pi}{12}) \sin^2(\frac{\pi}{12}) = 2(2-\sqrt{3})(\frac{1-\cos(\pi/6)}{2}) = (2-\sqrt{3})(1-\frac{\sqrt{3}}{2}) = \frac{(2-\sqrt{3})^2}{2} = \frac{7-4\sqrt{3}}{2}$.
Also,$\frac{(\sqrt{3}-1)^4}{8} = \frac{(4-2\sqrt{3})^2}{8} = \frac{28-16\sqrt{3}}{8} = \frac{7-4\sqrt{3}}{2}$. Thus,$(IV) ightarrow (P)$.
Therefore,$(I) ightarrow (Q), (II) ightarrow (T), (III) ightarrow (S), (IV) ightarrow (P)$.

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

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