The triangle $PQR$ of area $A$ is inscribed in the parabola $y^2 = 4ax$ such that the vertex $P$ lies at the vertex of the parabola and the base $QR$ is a focal chord. The modulus of the difference of the ordinates of the points $Q$ and $R$ 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?

If the curves $y^2 = 6x$ and $9x^2 + by^2 = 16$ intersect each other at right angles,then the value of $b$ is

The equation of the common tangent to the curves ${y^2} = 8x$ and $xy = -1$ is

If the line $y = \sqrt{3}x$ cuts the curve $x^4 + ax^2y + bxy + cx + dy + 6 = 0$ at $A$,$B$,$C$,and $D$,then the value of $OA \cdot OB \cdot OC \cdot OD$ is,(where $O$ is the origin).

The quadratic equation whose roots are $l$ and $m$,where $l = \lim_{\theta \rightarrow 0} \left( \frac{3 \sin \theta - 4 \sin^2 \theta}{\theta} \right)$ and $m = \lim_{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta(1 - \tan^2 \theta)}$,is: