The eccentricity of an ellipse $E$ with centre at the origin $O$ is $\frac{\sqrt{3}}{2}$ and its directrices are $x= \pm \frac{4 \sqrt{6}}{3}$. Let $\mathrm{H}: \frac{x^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}=1$ be a hyperbola whose eccentricity is equal to the length of semi-major axis of E , and whose length of latus rectum is equal to the length of minor axis of E . Then the distance between the foci of H is :

Let $e_1$ and $e_2$ be two distinct roots of the equation $x^2-a x+2=0$. Let the sets $\left\{a \in \mathbb{R}: e_1\right.$ and $e_2$ are the eccentricities of hyperbolas $\}=(\alpha, \beta)$, and $\left\{a \in \mathbb{R}: e_1\right.$ and $e_2$ are the eccentricities of an ellipse and a hyperbola, respectively $\}=(\gamma, \infty)$.

Then $\alpha^2+\beta^2+\gamma^2$ is equal to:

If the eccentricity $e$ of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, passing through $(6,4 \sqrt{3})$, satisfies $15\left(e^2+1\right)=34 e$, then the length of the latus rectum of the hyperbola $\frac{x^2}{b^2}-\frac{y^2}{2\left(a^2+1\right)}=1$ is:

Let the eccentricity e of a hyperbola satisfy the equation $6 \mathrm{e}^2-11 \mathrm{e}+3=0$. If the foci of the hyperbola are $(3,5)$ and $(3,-4)$, then the length of its latus rectum is :