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