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 |
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(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) $\frac{3 \sqrt{3}}{2}$ |
The correct option is:
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