Let the foci of the hyperbola coincide with the foci of the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$. If the eccentricity of the hyperbola is $5$,then the length of its latus rectum is:

If the equation of the hyperbola having $(8,3)$ and $(0,3)$ as foci and $\frac{4}{3}$ as eccentricity is $\frac{(x-\alpha)^2}{p}-\frac{(y-\beta)^2}{q}=1$,then $p+q=$

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:

Find the equation of the tangent to the hyperbola $\frac{x^2}{3} - \frac{y^2}{2} = 1$ which is equally inclined to the axes.

The midpoint of the chord $4x - 3y = 5$ of the hyperbola $2x^2 - 3y^2 = 12$ is

Let the eccentricity of the hyperbola $H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be $\sqrt{\frac{5}{2}}$ and the length of its latus rectum be $6\sqrt{2}$. If $y = 2x + c$ is a tangent to the hyperbola $H$,then the value of $c^2$ is equal to