The eccentricity of the hyperbola $\frac{x^2}{16} - \frac{y^2}{25} = 1$ is

The auxiliary circle of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by:

The locus of a point $P(\alpha, \beta)$ moving under the condition that the line $y = \alpha x + \beta$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is

If a hyperbola passes through the point $P(10, 16)$ and it has vertices at $(\pm 6, 0)$,then the equation of the normal to it at $P$ is

$A$ hyperbola with centre at $(0,0)$ has its transverse axis along the $X$-axis,and its length is $12$. If $(8,2)$ is a point on the 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: $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$.