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 equation of the hyperbola satisfying the give conditions: Vertices $(0,\,\pm 5),$ foci $(0,\,±8)$

Let $0 < \theta  < \frac{\pi }{2}$. If the eccentricity of the hyperbola $\frac{{{x^2}}}{{{{\cos }^2}\,\theta }} - \frac{{{y^2}}}{{{{\sin }^2}\,\theta }} = 1$ is greater than $2$, then the length of its latus rectum lies in the interval

Let $H : \frac{ x ^{2}}{ a ^{2}}-\frac{y^{2}}{ b ^{2}}=1$, a $>0, b >0$, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is $4(2 \sqrt{2}+\sqrt{14})$. If the eccentricity $H$ is $\frac{\sqrt{11}}{2}$, then value of $a^{2}+b^{2}$ is equal to

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $16 x^{2}-9 y^{2}=576$