If the foci of the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{{b^2}}} = 1$ coincide with the foci of the hyperbola $\frac{{{x^2}}}{{144}} - \frac{{{y^2}}}{{81}} = \frac{1}{{25}},$ then $b^2$ is equal to

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 $x^2-3 y^2=3$

The line $lx + my + n = 0$ will be a tangent to the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$, if

The locus of a point $P(\alpha ,\,\beta )$ moving under the condition that the line $y = \alpha x + \beta $ is a tangent to the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ is

 Find the equation of the hyperbola where foci are $(0,\,±12)$ and the length of the latus rectum is $36.$

For $0<\theta<\pi / 2$, if the eccentricity of the hyperbola $\mathrm{x}^2-\mathrm{y}^2 \operatorname{cosec}^2 \theta=5$ is $\sqrt{7}$ times eccentricity of the ellipse $x^2 \operatorname{cosec}^2 \theta+y^2=5$, then the value of $\theta$ is :