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 $\varphi$ with the positive $x$-axis.

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

The correct option is:

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

Answer 1

Answer: (C)

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