What is the equation of the ellipse with foci $(\pm 2, 0)$ and eccentricity $e = \frac{1}{2}$?

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 $FGH$ is	$(P) \frac{(\sqrt{3}-1)^4}{8}$
$(II)$ If $\phi=\frac{\pi}{3}$,then the area of the triangle $FGH$ is	$(Q) 1$
$(III)$ If $\phi=\frac{\pi}{6}$,then the area of the triangle $FGH$ is	$(R) \frac{3}{4}$
$(IV)$ If $\phi=\frac{\pi}{12}$,then the area of the triangle $FGH$ is	$(S) \frac{1}{2\sqrt{3}}$
	$(T) \frac{3\sqrt{3}}{2}$

The correct option is:

For the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$,a chord $PQ$ subtends a right angle at the center. What is the locus of the point of intersection of the tangents at $P$ and $Q$?

The product of the perpendiculars drawn from the points $(\pm \sqrt{a^2 - b^2}, 0)$ to the line $\frac{x}{a}\cos \theta + \frac{y}{b}\sin \theta = 1$ is:

If tangents are drawn to the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$ at the ends of the latus recta,then the area of the quadrilateral thus formed is

Let $E$ be an ellipse whose major axis is the $X$-axis and minor axis is the $Y$-axis. If the distances of a point $P \left(\frac{5}{2}, 2 \sqrt{3}\right)$ on $E$ from its foci are $\frac{7}{2}$ and $\frac{13}{2}$,then the eccentricity of the ellipse $E$ is