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

Let the ellipse $E_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$ and $E_2: \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1, A < B$ have the same eccentricity $e = \frac{1}{\sqrt{3}}$. Let the product of their lengths of latus rectums be $\frac{32}{\sqrt{3}}$,and the distance between the foci of $E_1$ be $4$. If $E_1$ and $E_2$ meet at $A, B, C$ and $D$,then the area of the quadrilateral $ABCD$ equals:

Let the product of the focal distances of the point $\left(\sqrt{3}, \frac{1}{2}\right)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ be $\frac{7}{4}$. Then the absolute difference of the eccentricities of two such ellipses is

An ellipse passing through $(4 \sqrt{2}, 2 \sqrt{6})$ has foci at $(-4, 0)$ and $(4, 0)$. Then,its eccentricity is

The length of the minor axis (along $y$-axis) of an ellipse in the standard form is $\frac{4}{\sqrt{3}}$. If this ellipse touches the line $x+6y=8$,then its eccentricity is

The equation of the ellipse whose centre is $(2, -3)$,one of the foci is $(3, -3)$ and the corresponding vertex is $(4, -3)$ is