If $\alpha, \beta$ are the eccentric angles of the extremities of a focal chord (other than the major axis) of the ellipse $x^2+4y^2=4$,then $\sqrt{3} \cos \frac{\alpha+\beta}{2} =$

The equations of the directrices of the ellipse $16x^2 + 25y^2 = 400$ are

In an ellipse,if the distance between the foci is $6$ units and the length of its minor axis is $8$ units,then its eccentricity 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:

The eccentric angle of a point on the ellipse $x^2 + 3y^2 = 6$ at a distance of $2$ units from the centre of the ellipse is:

If the midpoint of a chord of the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ is $(\sqrt{2}, 4/3)$,and the length of the chord is $\frac{2 \sqrt{\alpha}}{3}$,then $\alpha$ is :