$A$ conic passes through the point $(2, 4)$ and is such that the segment of any of its tangents at any point contained between the coordinate axes is bisected at the point of tangency. Then the foci of the conic are:

For $0 < \theta < \pi / 2$,if the eccentricity of the hyperbola $x^2 - y^2 \operatorname{cosec}^2 \theta = 5$ is $\sqrt{7}$ times the eccentricity of the ellipse $x^2 \operatorname{cosec}^2 \theta + y^2 = 5$,then the value of $\theta$ is:

The locus of the foot of the perpendicular drawn from the centre upon any tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is

The value of $b^2$ such that the foci of the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$ and the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ coincide is

The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(b>a)$ and the parabola $y^2=8ax$ intersect at right angles. If $e$ is the eccentricity of the ellipse,then $e^4$ is equal to

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?