The centre of an ellipse is $C$,$PN$ is any ordinate,and $A$,$A'$ are the end points of the major axis. Then the value of $\frac{PN^2}{AN \cdot A'N}$ is

The ellipse having its foci $(0, \pm 1)$ and major axis of length $\sqrt{5}$ is

The eccentricity of the ellipse $4x^2 + 9y^2 + 8x + 36y + 4 = 0$ is

If tangents are drawn to the ellipse $x^2+2y^2=2$,then the locus of the mid-points of the intercepts made by those tangents between the coordinate axes 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:

If the normal drawn at the point $(2, -1)$ to the ellipse $x^2 + 4y^2 = 8$ meets the ellipse again at $(a, b)$,then $17a =$