Find the equation for the ellipse that satisfies the given conditions: Vertices $(\pm 6,\,0),$ foci $(\pm 4,\,0)$

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 $........$

Let the line $y=m x$ and the ellipse $2 x^{2}+y^{2}=1$ intersect at a ponit $\mathrm{P}$ in the first quadrant. If the normal to this ellipse at $P$ meets the co-ordinate axes at $\left(-\frac{1}{3 \sqrt{2}}, 0\right)$ and $(0, \beta),$ then $\beta$ is equal to

A tangent is drawn to the ellipse $\frac{{{x^2}}}{{32}} + \frac{{{y^2}}}{8} = 1$ from the point $A(8, 0)$ to touch the ellipse at point $P.$ If the normal at $P$ meets the major axis of ellipse at point $B,$ then the length $BC$ is equal to (where $C$ is centre of ellipse) - ............ $\mathrm{units}$

Let $P$ be a variable point on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ with foci ${F_1}$ and ${F_2}$. If $A$ is the area of the triangle $P{F_1}{F_2}$, then maximum value of $A$ is

In an ellipse $9{x^2} + 5{y^2} = 45$, the distance between the foci is