The eccentricity of an ellipse is $2/3$,the length of the latus rectum is $5$,and the centre is $(0, 0)$. The equation of the ellipse is:

The centre of a circle $C$ is at the centre of the ellipse $E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$. Let $C$ pass through the foci $F_1$ and $F_2$ of $E$ such that the circle $C$ and the ellipse $E$ intersect at four points. Let $P$ be one of these four points. If the area of the triangle $PF_1F_2$ is $30$ and the length of the major axis of $E$ is $17$,then the distance between the foci of $E$ is:

$A$ line passing through the point $P(\sqrt{5}, \sqrt{5})$ intersects the ellipse $\frac{x^2}{36} + \frac{y^2}{25} = 1$ at $A$ and $B$ such that $(PA) \cdot (PB)$ is maximum. Then $5(PA^2 + PB^2)$ is equal to:

The tangent at point $(a \cos \theta, b \sin \theta)$, where $0 < \theta < \frac{\pi}{2}$, to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ meets the $x$-axis at $T$ and the $y$-axis at $T_1$. Then the value of $\min_{0 < \theta < \frac{\pi}{2}} (OT)(OT_1)$ is

If $S$ and $S'$ are the two foci of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a < b$,and $P(x_1, y_1)$ is a point on the ellipse,then $SP + S'P = \dots$

If the centre,one of the foci,and the semi-major axis of an ellipse are $(0, 0)$,$(0, 3)$,and $5$ respectively,then its equation is: