The minimum area of a triangle formed by any tangent to the ellipse $\frac{x^2}{16} + \frac{y^2}{81} = 1$ and the coordinate axes is

The centre of the ellipse $\frac{(x+y-3)^2}{9}+\frac{(x-y+1)^2}{16}=1$ is

On the ellipse $\frac{x^{2}}{8}+\frac{y^{2}}{4}=1$,let $P$ be a point in the second quadrant such that the tangent at $P$ to the ellipse is perpendicular to the line $x+2y=0$. Let $S$ and $S'$ be the foci of the ellipse and $e$ be its eccentricity. If $A$ is the area of the triangle $SPS'$,then the value of $(5-e^{2}) \cdot A$ is:

If the tangents on the ellipse $4x^2 + y^2 = 8$ at the points $(1, 2)$ and $(a, b)$ are perpendicular to each other,then $a^2$ is equal to

If an ellipse with its axes as coordinate axes,$2a$ and $2b$ as the lengths of its major and minor axes respectively,passes through the points $(2,2)$ and $(3,1)$,then $3a^2+5b^2=$

$A$ point moves such that the sum of its distances from $(ae, 0)$ and $(-ae, 0)$ is $2a$. Then the equation to its locus,where $b^2 = a^2(1 - e^2)$,is