Let the equation of an ellipse be $\frac{x^{2}}{144}+\frac{y^{2}}{25}=1$. Then, the radius of the circle with center $(0, \sqrt{2})$ and passing through the foci of the ellipse is

The coordinate of a point on the auxiliary circle of the ellipse $x^{2}+2y^{2}=4$ corresponding to the point on the ellipse whose eccentric angle is $60^{\circ}$ will be

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

The eccentricity of the ellipse $9x^2 + 5y^2 - 18x - 20y - 16 = 0$ is:

$A$ rod $AB$ of length $15 \ cm$ rests between two coordinate axes such that the end point $A$ lies on the $x-$axis and the end point $B$ lies on the $y-$axis. $A$ point $P(x, y)$ is taken on the rod such that $AP = 6 \ cm$. Show that the locus of $P$ is an ellipse.

The length of the latus rectum of an ellipse is $\frac{18}{5}$ and eccentricity is $\frac{4}{5}$,then the equation of the ellipse is...