Let $\frac{x^2}{f\left(a^2+7 a+3\right)}+\frac{y^2}{f(3 a+15)}=1$ represent an ellipse with major axis along $y$-axis, where $f$ is a strictly decreasing positive function on $\mathbf{R}$. If the set of all possible values of $a$ is $\mathbf{R}-[\alpha, \beta]$, then $\alpha^2+\beta^2$ is equal to :

Let $x=9$ be a directrix of an ellipse E , whose centre is at the origin and eccentricity is $\frac{1}{3}$. Let $\mathrm{P}(\alpha, 0)$, $\alpha>0$, be a focus of E and AB be a chord passing through P . Then the locus of the mid point of AB is :

Let a focus of the ellipse $\mathrm{E}: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ be $\mathrm{S}(4,0)$ and its eccentricity be $\frac{4}{5}$. If the point $\mathrm{P}(3, \alpha)$ lies on E and O is the origin, then the area of $\triangle \mathrm{POS}$ is equal to:

Let $\mathrm{P}(3 \cos \alpha, 2 \sin \alpha), \alpha \neq 0$, be a point on the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1, \mathrm{Q}$ be a point on the circle $x^2+y^2-14 x-14 y+82=0$ and R be a point on the line $x+y=5$ such that the centroid of the triangle PQR is $\left(2+\cos \alpha, 3+\frac{2}{3} \sin \alpha\right)$. Then the sum of the ordinates of all possible points R is: