If the product of eccentricities of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=-1$ is $1$,then $b^2=$

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 $e_{1}$ and $e_{2}$ are the eccentricities of the ellipse $\frac{x^{2}}{18}+\frac{y^{2}}{4}=1$ and the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$ respectively,and $(e_{1}, e_{2})$ is a point on the ellipse $15x^{2}+3y^{2}=k$,then $k$ is equal to:

$A$ tangent is drawn to the parabola $y^2 = 4x$ at the point $P(t^2, 2t)$,where the abscissa $t^2$ lies in the interval $[1, 4]$. The maximum possible area of the triangle formed by the tangent at $P$,the ordinate of the point $P$,and the $x$-axis is equal to

Find the area of the region enclosed by the equation $2|x| + 3|y| = 6$ in square units.

The slopes of the common tangents to the parabola $(x - 1)^2 = 4(y - 2)$ and the ellipse $\frac{(x - 1)^2}{1} + \frac{(y - 2)^2}{2} = 1$ are $m_1$ and $m_2$. Then,$m_1^2 + m_2^2$ is equal to: