Coordinate Geometry: JEE Advanced 2022 Mathematics question

Consider the ellipse \[ \frac{x²}{4}+\frac{y²}{3}=1 \] 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 (I) If $\phi=\frac{\pi}{4}$, then the area of the triangle $F G H$ is (II) If $\phi=\frac{\pi}{3}$, then the area of the triangle $F G H$ is (III) If $\phi=\frac{\pi}{6}$, then the area of the triangle $F G H$ is (IV) If $\phi=\frac{\pi}{12}$, then the area of the triangle $F G H$ is List-II (P) $\frac{(\sqrt{3}-1)^4}{8}$ (Q) 1 (R) $\frac{3}{4}$ (S) $\frac{1}{2 \sqrt{3}}$ (T) $\frac{3 \sqrt{3}}{2}$ The correct option is: