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 :

Let $\mathrm{H}: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be a hyperbola such that the distance between its foci is 6 and the distance between its directrices is $\frac{8}{3}$. If the line $x=\alpha$ intersects the hyperbola H at the points A and B such that the area of the triangle AOB is $4 \sqrt{15}$, where O is the origin, then $\alpha^2$ equals