The locus of the midpoints of the segments of the tangents to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ intercepted between the coordinate axes is:

Find the equation of the ellipse whose vertices are $(\pm 13, 0)$ and foci are $(\pm 5, 0)$.

$P(\theta_1)$ and $Q(\theta_2)$ are two points on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with eccentricity $e$. If $PSQ$ is a focal chord and $\tan \left(\frac{\theta_1}{2}\right) \tan \left(\frac{\theta_2}{2}\right)=-(2 \sqrt{2}+3)$,then $e$ and $S$ are

If the points of intersection of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$ and the circle $x^{2}+y^{2}=4b$ (where $b > 4$) lie on the curve $y^{2}=3x^{2}$,then $b$ is equal to:

If the chord through the points whose eccentric angles are $\theta$ and $\phi$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ passes through the focus,then the value of $(1 + e) \tan(\frac{\theta}{2}) \tan(\frac{\phi}{2})$ is

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: