The length of the latus rectum of the ellipse $16x^2 + 25y^2 = 400$ is

If the radius of the largest circle with centre $(2,0)$ inscribed in the ellipse $x^2+4y^2=36$ is $r$,then $12r^2$ is equal to

If $S$ and $S'$ are the two foci of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a < b$,and $P(x_1, y_1)$ is a point on the ellipse,then $SP + S'P = \dots$

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:

If the distance between the foci of an ellipse is half the length of its latus rectum,then the eccentricity of the ellipse is

If $P(x, y)$,$F_1 = (3, 0)$,$F_2 = (-3, 0)$ and $16x^2 + 25y^2 = 400$,then $PF_1 + PF_2 = \dots$