If $3 \sqrt{2} x - 4 y = 12$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $\frac{5}{4}$ is its eccentricity,then $a^2 - b^2 =$

The eccentricity of a hyperbola passing through the points $(3, 0)$ and $(3\sqrt{2}, 2)$ is:

Let the foci and length of the latus rectum of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ be $(\pm 5, 0)$ and $\sqrt{50}$,respectively. Then,the square of the eccentricity of the hyperbola $\frac{x^2}{b^2}-\frac{y^2}{a^2b^2}=1$ equals

The equation of a hyperbola,whose foci are $(5, 0)$ and $(-5, 0)$ and the length of whose conjugate axis is $8$,is

For the hyperbola $x^2 \sec^2 \theta - y^2 \csc^2 \theta = 1$,which of the following remains constant when $\theta$ varies?

What is the area of the triangle formed by the lines $x - y = 0$,$x + y = 0$,and any tangent to the hyperbola $x^{2} - y^{2} = a^{2}$?