If $(4, 0)$ and $(-4, 0)$ are the vertices and $(6, 0)$ and $(-6, 0)$ are the foci of a hyperbola,then its eccentricity is

Find the coordinates of the foci and the vertices,the eccentricity,and the length of the latus rectum of the hyperbola $9 y^{2}-4 x^{2}=36$.

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

$PQ$ is a double ordinate of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ such that $\triangle OPQ$ is an equilateral triangle, where $O$ is the centre of the hyperbola. Then the eccentricity $e$ of the hyperbola satisfies:

The equation ${x^2} - 16xy - 11{y^2} - 12x + 6y + 21 = 0$ represents

The equation of the hyperbola whose foci are the same as the foci of the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ and whose eccentricity is $2$ is: