In the ellipse, minor axis is $8$ and eccentricity is $\frac{{\sqrt 5 }}{3}$. Then major axis is

If end points of latus rectum of an ellipse are vertices of a square, then eccentricity of ellipse will be -

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{36}+\frac{y^2} {16}=1$

Find the equation for the ellipse that satisfies the given conditions : Vertices $(\pm 5,\,0),$ foci $(\pm 4,\,0)$

Let $E$ be the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$. For any three distinct points $P, Q$ and $Q^{\prime}$ on $E$, let $M(P, Q)$ be the mid-point of the line segment joining $P$ and $Q$, and $M \left( P , Q ^{\prime}\right)$ be the mid-point of the line segment joining $P$ and $Q ^{\prime}$. Then the maximum possible value of the distance between $M ( P , Q )$ and $M \left( P , Q ^{\prime}\right)$, as $P, Q$ and $Q^{\prime}$ vary on $E$, is. . . . .

Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $\sqrt{14}$. Then the square of the eccentricity of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is :