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 = $

Find the equation for the ellipse that satisfies the given conditions: $b=3, c=4,$ centre at the origin; foci on the $x$-axis.

The locus of the midpoint of the line segment joining the point $(4,3)$ and the points on the ellipse $x^{2}+2y^{2}=4$ is an ellipse with eccentricity:

In an ellipse with its centre at the origin,if the difference between the lengths of the major axis and the minor axis is $10$ and one of the foci is at $(0, 5\sqrt{3})$,then the length of its latus rectum 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:

$S^{\prime}$ is the focus of the ellipse $\frac{x^2}{25}+\frac{y^2}{b^2}=1, (b < 5)$ lying on the negative $X$-axis and $P(\theta)$ is a point on this ellipse. If the distance between the foci of this ellipse is $8$ and $S^{\prime}P = 7$,then $\theta =$