Find the equation of the hyperbola satisfying the given conditions: Vertices $(0, \pm 3)$,foci $(0, \pm 5)$.

Let $H: \frac{-x^2}{a^2}+\frac{y^2}{b^2}=1$ be the hyperbola,whose eccentricity is $\sqrt{3}$ and the length of the latus rectum is $4\sqrt{3}$. Suppose the point $(\alpha, 6)$,where $\alpha > 0$,lies on $H$. If $\beta$ is the product of the focal distances of the point $(\alpha, 6)$,then $\alpha^2+\beta$ is equal to:

If the product of the perpendicular distances from any point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ to its asymptotes is $\frac{36}{13}$ and its eccentricity is $\frac{\sqrt{13}}{3}$,then $a - b =$

If the distance between the foci of a hyperbola $H$ is $26$ and the distance between its directrices is $\frac{50}{13}$,then the eccentricity of the conjugate hyperbola of the hyperbola $H$ is

If the equation of the tangent drawn at $(h, k)$ to the hyperbola $\frac{(x-1)^2}{1}-\frac{(y-2)^2}{2}=1$ is $x=2$,then $h+k=$

Let the foci of a hyperbola be $(1, 14)$ and $(1, -12)$. If it passes through the point $(1, 6)$,then the length of its latus-rectum is: