If $S$ and $S^{\prime}$ are the foci of the ellipse $\frac{x^2}{18}+\frac{y^2}{9}=1$ and $P$ is a point on the ellipse,then $\min \left(SP \cdot S^{\prime}P\right) + \max \left(SP \cdot S^{\prime}P\right)$ is equal to:

The equation of the ellipse whose one focus is at $(4, 0)$ and whose eccentricity is $4/5$,is

If $F_1$ and $F_2$ are the foci of the ellipse $16 x^2+25 y^2=400$ and $P$ is any point on it,then the value of the product $P F_1 \cdot P F_2$ lies in the interval

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 equation $2x^2 + 3y^2 = 30$ represents

If the product of the lengths of the perpendiculars drawn from the foci to the tangent $y = \frac{-3}{4}x + 3\sqrt{2}$ of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $9$,then the eccentricity of that ellipse is