If $4x + 2y + n = 0$ is a normal to the ellipse $\frac{x^2}{36} + \frac{y^2}{16} = 1$,then $n = $

The distance between the foci of an ellipse is $16$ and its eccentricity is $\frac{1}{2}$. The length of the major axis of the ellipse is:

If the eccentricity of an ellipse is $5/8$ and the distance between its foci is $10$,then its latus rectum is

Let the length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b)$ be $30$. If its eccentricity is the maximum value of the function $f(t)=-\frac{3}{4}+2t-t^{2}$, then $(a^{2}+b^{2})$ is equal to -

The product of the lengths of the perpendiculars drawn from the two foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$ to the tangent at any point on the ellipse is

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 $\qquad$