If $\theta$ is the eccentric angle of a point on the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ such that the distance of the point from the center is $5$, then $\theta = ......$

The sides of the rectangle of greatest area that can be inscribed in the ellipse $x^2+4y^2=64$ are :

Equations of the latus rectum of the ellipse $9x^2+4y^2-18x-8y-23=0$ are:

The equations of two ellipses are $\frac{x^2}{4}+\frac{y^2}{2}=1$ and $\frac{x^2}{36}+\frac{y^2}{b^2}=1$. If the product of their eccentricities is $\frac{\sqrt{2}}{3}$,then the product of the length of the major axis and minor axis of the second ellipse is $\qquad$

On the ellipse $4x^2 + 9y^2 = 1$,the points at which the tangents are parallel to the line $8x = 9y$ are

The product of the perpendicular distances drawn from the points $(3,0)$ and $(-3,0)$ to the tangent of the ellipse $\frac{x^2}{36}+\frac{y^2}{27}=1$ at the point $\left(3, \frac{9}{2}\right)$ is: