Question

Consider the ellipse
$\frac{x^2}{4}+\frac{y^2}{3}=1 .$
Let $H (\alpha, 0), 0< \alpha< 2$, be a point. A straight line drawn through $H$ parallel to the $y$-axis crosses the ellipse and its auxiliary circle at points $E$ and $F$ respectively, in the first quadrant. The tangent to the ellipse at the point $E$ intersects the positive $x$-axis at a point $G$. Suppose the straight line joining $F$ and the origin makes an angle $\phi$ with the positive $x$-axis.

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

The correct option is:

• A. (I) $\rightarrow$ (R); (II) $\rightarrow$ (S); (III) $\rightarrow$ (Q); (IV) $\rightarrow$ (P)
• B. (I) $\rightarrow$ (R); (II) $\rightarrow$ (T); (III) $\rightarrow$ (S); (IV) $\rightarrow$ (P)
• C. (I) $\rightarrow$ (Q); (II) $\rightarrow$ (T); (III) $\rightarrow$ (S); (IV) $\rightarrow$ (P)
• D. (I) $\rightarrow$ (Q); (II) $\rightarrow$ (S); (III) $\rightarrow$ (Q); (IV) $\rightarrow$ (P)

Answer 1

Answer: (C)

Let $F (2 \cos \phi, 2 \sin \phi)$
$\& E (2 \cos \phi, \sqrt{3} \sin \phi)$
EG : $\frac{x}{2} \cos \phi+\frac{y}{\sqrt{3}} \sin \phi=1$
$\therefore G \left(\frac{2}{\cos \phi}, 0\right)$ and $\alpha=2 \cos \phi$
$\operatorname{ar}(\Delta FGH )=\frac{1}{2} HG \times FH$
$=\frac{1}{2}\left(\frac{2}{\cos \phi}-2 \cos \phi\right) \times 2 \sin \phi$
$f(\phi)=2 \tan \phi \sin ^2 \phi$
$\therefore$ (I) $f \left(\frac{\pi}{4}\right)=1$
(II) $f \left(\frac{\pi}{3}\right)=\frac{3 \sqrt{3}}{2}$
(III) $f \left(\frac{\pi}{6}\right)=\frac{1}{2 \sqrt{3}}$
(IV) $f\left(\frac{\pi}{12}\right)=2(2-\sqrt{3})\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)^2=(4-2 \sqrt{3}) \frac{(\sqrt{3}-1)^2}{8}=\frac{(\sqrt{3}-1)^4}{8}$
$\therefore( I ) \rightarrow( Q ) ;( II ) \rightarrow( T ) ;( III ) \rightarrow( S ) ;( IV ) \rightarrow( P )$