If the product of the perpendicular distances from any point on the hyperbola $\frac{{{x^2}}}{{{a^2}}}\,\, - \,\,\frac{{{y^2}}}{{{b^2}}}\,\,\, = \,1$ of eccentricity $e =\sqrt 3 \,$ from its asymptotes is equal to $6$, then the length of the transverse axis of the hyperbola is

If the foci of a hyperbola are same as that of the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$ and the eccentricity of the hyperbola is $\frac{15}{8}$ times the eccentricity of the ellipse, then the smaller focal distance of the point $\left(\sqrt{2}, \frac{14}{3} \sqrt{\frac{2}{5}}\right)$ on the hyperbola, is equal to

Find the coordinates of the foci and the vertices, the eccentricity,the length of the latus rectum of the hyperbolas : $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1.$

The locus of the middle points of the chords of hyperbola $3{x^2} - 2{y^2} + 4x - 6y = 0$ parallel to $y = 2x$ 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

The magnitude of the gradient of the tangent at an extremity of latera recta of the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ is equal to (where $e$ is the eccentricity of the hyperbola)