Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse, $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$ from any of its foci?

The length of the axes of the conic $9{x^2} + 4{y^2} - 6x + 4y + 1 = 0$, are

The line $x =8$ is the directrix of the ellipse $E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with the corresponding focus $(2,0)$. If the tangent to $E$ at the point $P$ in the first quadrant passes through the point $(0,4 \sqrt{3})$ and intersects the $x$-axis at $Q$, then $(3PQ)^2$ is equal to $........$

The eccentricity of the ellipse ${\left( {\frac{{x - 3}}{y}} \right)^2} + {\left( {1 - \frac{4}{y}} \right)^2} = \frac{1}{9}$ is

A vertical line passing through the point $(h, 0)$ intersects the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at the points $P$ and $Q$. Let the tangents to the ellipse at $P$ and $Q$ meet at the point $R$. If $\Delta(h)=$ area of the triangle $P Q R, \Delta_1=\max _{1 / 2 \leq h \leq 1} \Delta(h)$ and $\Delta_2=\min _{1 / 2 \leq h \leq 1} \Delta(h)$, then $\frac{8}{\sqrt{5}} \Delta_1-8 \Delta_2=$

Minimum area of the triangle by any tangent to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ with the coordinate axes is