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

Let O be the origin, and P and Q be two points on the rectangular hyperbola $xy = 12$ such that the midpoint of the line segment PQ is $\left( \frac{1}{2}, -\frac{1}{2} \right)$. Then the area of the triangle OPQ equals :

Let the ellipse $E: \frac{x^2}{144} + \frac{y^2}{169} = 1$ and the hyperbola $H: \frac{x^2}{16} - \frac{y^2}{\lambda^2} = -1$ have the same foci. If $e$ and $L$

respectively denote the eccentricity and the length of the latus rectum of $H$, then the value of $24(e+L)$ is :