Length of latus rectum of hyperbola $\frac{{{x^2}}}{{{{\cos }^2}\alpha }} - \frac{{{y^2}}}{{{{\sin }^2}\alpha }} = 4\,is\,\left( {\alpha  \ne \frac{{n\pi }}{2},n \in I} \right)$

The equation of the hyperbola whose foci are $(6, 4)$ and $(-4, 4)$ and eccentricity $2$ is given by

Let the tangent drawn to the parabola $y ^{2}=24 x$ at the point $(\alpha, \beta)$ is perpendicular to the line $2 x$ $+2 y=5$. Then the normal to the hyperbola $\frac{x^{2}}{\alpha^{2}}-\frac{y^{2}}{\beta^{2}}=1$ at the point $(\alpha+4, \beta+4)$ does $NOT$ pass through the point.

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $5 y^{2}-9 x^{2}=36$

If the eccentricity of the hyperbola $x^2 - y^2 \sec^2 \alpha = 5$  is $\sqrt 3 $  times the eccentricity of the ellipse $x^2 \sec^2 \alpha + y^2 = 25, $ then a value of $\alpha$  is :

The tangent at an extremity (in the first quadrant) of latus rectum of the hyperbola $\frac{{{x^2}}}{4} - \frac{{{y^2}}}{5} = 1$ , meet $x-$ axis and $y-$ axis at $A$ and $B$ respectively. Then $(OA)^2 - (OB)^2$ , where $O$ is the origin, equals