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

$A$ chord $PQ$ of the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ subtends a right angle at its centre. The locus of the point of intersection of tangents drawn at $P$ and $Q$ is-

Let the length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b)$ be $30$. If its eccentricity is the maximum value of the function $f(t)=-\frac{3}{4}+2t-t^{2}$, then $(a^{2}+b^{2})$ is equal to -

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

Let each of the two ellipses $E_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, (a > b)$ and $E_2: \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1, (A < B)$ have eccentricity $\frac{4}{5}$. Let the lengths of the latus recta of $E_1$ and $E_2$ be $\ell_1$ and $\ell_2$, respectively, such that $2\ell_1^2 = 9\ell_2$. If the distance between the foci of $E_1$ is $8$, then the distance between the foci of $E_2$ is:

If the tangents drawn from a point $P$ to the ellipse $4 x^2+9 y^2-24 x+36 y=0$ are perpendicular,then the locus of $P$ is