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
(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:

Let S be the circle in the XY-plane defined the equation x² + y² = 4.

Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect the coordinate axes at the points M and N. Then, the mid-point of the line segment MN must lie on the curve

Let $${F_1}\left( {{x_1},0} \right)$$ and $${F_2}\left( {{x_2},0} \right)$$ for $${{x_1} < 0}$$ and $${{x_2} > 0}$$, be the foci of the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 8} = 1$$. Suppose a parabola having vertex at the origin and focus at $${F_2}$$ intersects the ellipse at point $$M$$ in the first quadrant and at point $$N$$ in the fourth quadrant.

The orthocentre of the triangle $${F_1}MN$$ is

Let $${F_1}\left( {{x_1},0} \right)$$ and $${F_2}\left( {{x_2},0} \right)$$ for $${{x_1} < 0}$$ and $${{x_2} > 0}$$, be the foci of the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 8} = 1$$. Suppose a parabola having vertex at the origin and focus at $${F_2}$$ intersects the ellipse at point $$M$$ in the first quadrant and at point $$N$$ in the fourth quadrant.

If the tangents to the ellipse at $$M$$ and $$N$$ meet at $$R$$ and the normal to the parabola at $$M$$ meets the $$x$$-axis at $$Q$$, then the ratio of area of the triangle $$MQR$$ to area of the quadrilateral $$M{F_1}N{F_2}$$is