The values that $m$ can take,so that the straight line $y=4x+m$ touches the curve $x^2+4y^2=4$ are

The eccentricity of the ellipse $(x - 3)^2 + (y - 4)^2 = \frac{y^2}{9}$ is

The locus of the point $P(x, y)$ satisfying the relation $\sqrt{(x - 3)^2 + (y - 1)^2} + \sqrt{(x + 3)^2 + (y - 1)^2} = 6$ is

Let the length of a latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be $10$. If its eccentricity is the minimum value of the function $f(t) = t^2 + t + \frac{11}{12}$,$t \in R$,then $a^2 + b^2$ is equal to:

Statement $-1$: If two tangents are drawn to an ellipse from a single point and if they are perpendicular to each other,then the locus of that point is always a circle.
Statement $-2$: For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,the locus of the point from which two perpendicular tangents are drawn is $x^2 + y^2 = a^2 + b^2$.

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: