Let the equations of two ellipses be $E_1: \frac{x^2}{3} + \frac{y^2}{2} = 1$ and $E_2: \frac{x^2}{16} + \frac{y^2}{b^2} = 1$. If the product of their eccentricities is $\frac{1}{2}$,then the length of the minor axis of ellipse $E_2$ is

The parametric equations of the ellipse whose foci are $(-3, 0)$ and $(9, 0)$ and eccentricity is $\frac{1}{3}$,are

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 chord of the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ having $(1,1)$ as its midpoint is $x+\alpha y=\beta$,then

If the line $y = 4x + c$ is a tangent to the ellipse $\frac{x^2}{8} + \frac{y^2}{4} = 1$,then $c = \dots$

The locus of the midpoint of the line segment joining the point $(4,3)$ and the points on the ellipse $x^{2}+2y^{2}=4$ is an ellipse with eccentricity: