The equation of the ellipse with directrix $3x+4y-5=0$,focus $(1,2)$ and eccentricity $e = \frac{1}{2}$,is

Let $A=\{(\alpha, \beta) \in R \times R :|\alpha-1| \leq 4 \text{ and }|\beta-5| \leq 6\}$ and $B=\left\{(\alpha, \beta) \in R \times R : 16(\alpha-2)^2+9(\beta-6)^2 \leq 144\right\}$. Then

The eccentricity of the ellipse $25x^2 + 16y^2 - 150x - 175 = 0$ is

If $P$ is any point on the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ and $S, S^{\prime}$ are its foci,then the maximum area (in sq. units) of $\Delta SPS^{\prime} =$

If the length of the major axis of an ellipse is three times the length of its minor axis,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: