For $\alpha$ belonging to an interval of length $\beta$,suppose $(\alpha, -\alpha)$ is an interior point of the ellipse $4x^2 + 5y^2 = 1$. Then,$(6\beta - 4)^{201} + 201 = $

If the distance between the foci of an ellipse is equal to the length of the latus rectum, then its eccentricity is

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:

Find the area of the quadrilateral formed by the tangents at the endpoints of the latus rectum of the ellipse $x^2 + 2y^2 = 2$.

If a tangent having slope $m = \frac{1}{3}$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ is a normal to the circle $(x + 1)^2 + (y + 1)^2 = 1$,then $a^2$ lies in the interval: