The directrix of the hyperbola is $\frac{{{x^2}}}{9} - \frac{{{y^2}}}{4} = 1$

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 the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be reciprocal to that of the ellips $x^2+4 y^2=4$. If the hyperbola passes through a focus of the ellipse, then

$(A)$ the equation of the hyperbola is $\frac{x^2}{3}-\frac{y^2}{2}=1$

$(B)$ a focus of the hyperbola is $(2,0)$

$(C)$ the eccentricity of the hyperbola is $\sqrt{\frac{5}{3}}$

$(D)$ the equation of the hyperbola is $x^2-3 y^2=3$

Let $\mathrm{P}$ be a point on the hyperbola $\mathrm{H}: \frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{4}=1$, in the first quadrant such that the area of triangle formed by $\mathrm{P}$ and the two foci of $\mathrm{H}$ is $2 \sqrt{13}$. Then, the square of the distance of $\mathrm{P}$ from the origin is

The equation of the hyperbola whose foci are the foci of the ellipse $\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{9} = 1$ and the eccentricity is $2$, is

If $e$ and $e’$ are eccentricities of hyperbola and its conjugate respectively, then