Let $f$ be a strictly decreasing function defined on $\mathbb{R}$ such that $f(x) > 0, \forall x \in \mathbb{R}$. Let $\frac{x^2}{f(a^2+5a+3)} + \frac{y^2}{f(a+15)} = 1$ be an ellipse with the major axis along the $y$-axis. The value of $a$ can lie in the interval$(s)$:

Let $S$ and $S^{\prime}$ be the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ and $P(\alpha, \beta)$ be a point on the ellipse in the first quadrant. If $(SP)^2+(S^{\prime}P)^2-SP \cdot S^{\prime}P=37$, then $\alpha^2+\beta^2$ is equal to:

Let the ellipse $E_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$ and $E_2: \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1, A < B$ have the same eccentricity $e = \frac{1}{\sqrt{3}}$. Let the product of their lengths of latus rectums be $\frac{32}{\sqrt{3}}$,and the distance between the foci of $E_1$ be $4$. If $E_1$ and $E_2$ meet at $A, B, C$ and $D$,then the area of the quadrilateral $ABCD$ equals:

$A$ rectangle of maximum area is inscribed in an ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$. Then its dimensions are:

$P$ is any point on the ellipse $9x^2 + 36y^2 = 324$,whose foci are $S$ and $S'$. Then $SP + S'P$ equals

An ellipse having the coordinate axes as its axes and its major axis along the $Y$-axis,passes through the point $(-3, 1)$ and has eccentricity $e = \sqrt{\frac{2}{5}}$. Then its equation is: