Let $x^2=4ky, k>0$ be a parabola with vertex $O(0,0)$. Let $BC$ be its latus rectum. An ellipse with center $P$ on $BC$ touches the parabola at $O$,and cuts $BC$ at points $D$ and $E$ such that $BD=DE=EC$ ($B, D, E, C$ in that order). The eccentricity of the ellipse is

Columns $1, 2$ and $3$ contain conics,equations of tangents to the conics,and points of contact,respectively.

$Column 1$	$Column 2$	$Column 3$
$(I) x^2+y^2=a^2$	$(i) my=m^2x+a$	$(P) (a/m^2, 2a/m)$
$(II) x^2+a^2y^2=a^2$	$(ii) y=mx+a\sqrt{m^2+1}$	$(Q) (-ma/\sqrt{m^2+1}, a/\sqrt{m^2+1})$
$(III) y^2=4ax$	$(iii) y=mx+\sqrt{a^2m^2-1}$	$(R) (-a^2m/\sqrt{a^2m^2+1}, 1/\sqrt{a^2m^2+1})$
$(IV) x^2-a^2y^2=a^2$	$(iv) y=mx+\sqrt{a^2m^2+1}$	$(S) (-a^2m/\sqrt{a^2m^2-1}, -1/\sqrt{a^2m^2-1})$

$(1)$ The tangent to a suitable conic (Column $1$) at $(\sqrt{3}, 1/2)$ is $\sqrt{3}x+2y=4$. Which combination is correct?
$(2)$ If a tangent to a suitable conic (Column $1$) is $y=x+8$ and its point of contact is $(8, 16)$,which combination is correct?
$(3)$ For $a=\sqrt{2}$,if a tangent is drawn to a suitable conic (Column $1$) at $(-1, 1)$,which combination is correct?

The eccentricity of an ellipse $E$ with centre at the origin $O$ is $\frac{\sqrt{3}}{2}$ and its directrices are $x = \pm \frac{4\sqrt{6}}{3}$. Let $H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be a hyperbola whose eccentricity is equal to the length of semi-major axis of $E$, and whose length of latus rectum is equal to the length of minor axis of $E$. Then the distance between the foci of $H$ is :

If $PQ$ is a double ordinate of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ such that $\triangle OPQ$ is an equilateral triangle,where $O$ is the centre of the hyperbola,then the eccentricity $e$ of the hyperbola satisfies:

Find the equation of one of the common tangents to the parabola $y^2 = 4x$ and the ellipse $\frac{x^2}{4} + \frac{y^2}{3} = 1$.

Let the circle $C$ touch the line $x - y + 1 = 0$,have the centre on the positive $x$-axis,and cut off a chord of length $\frac{4}{\sqrt{13}}$ along the line $-3x + 2y = 1$. Let $H$ be the hyperbola $\frac{x^2}{\alpha^2} - \frac{y^2}{\beta^2} = 1$,whose one of the foci is the centre of $C$ and the length of the transverse axis is the diameter of $C$. Then $2\alpha^2 + 3\beta^2$ is equal to . . . . . .