If $\theta$ and $\phi$ are eccentric angles of the ends of a pair of conjugate diameters of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,then $\theta - \phi$ is equal to

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 $SP + S'P = \dots$

Let $P$ be any point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. If $S_1$ and $S_2$ are its foci,then the maximum area of $\Delta PS_1S_2$ is (in square units):

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:

If $\beta$ is one of the angles between the normals to the ellipse $x^2 + 3y^2 = 9$ at the points $(3\cos \theta, \sqrt{3} \sin \theta)$ and $(-3\sin \theta, \sqrt{3} \cos \theta)$,where $\theta \in (0, \pi/2)$,then $\frac{2 \cot \beta}{\sin 2\theta}$ is equal to

If tangents are drawn from the point $(2 + 13 \cos \theta, 3 + 13 \sin \theta)$ to the ellipse $\frac{(x-2)^2}{25} + \frac{(y-3)^2}{144} = 1$,then the angle between them is: