$A$ point on the curve $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$ is

The equation $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos \theta$ represents:

Let $a>0, b>0$. Let $e$ and $\ell$ respectively be the eccentricity and length of the latus rectum of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Let $e^{\prime}$ and $\ell^{\prime}$ respectively be the eccentricity and length of the latus rectum of its conjugate hyperbola. If $e^{2}=\frac{11}{14} \ell$ and $(e^{\prime})^{2}=\frac{11}{8} \ell^{\prime}$,then the value of $77a+44b$ is equal to

The equation of the normal at the point $(a \sec \theta, b \tan \theta)$ of the curve $b^2 x^2 - a^2 y^2 = a^2 b^2$ is

The equation of the hyperbola which passes through the point $(2,3)$ and has the asymptotes $4x+3y-7=0$ and $x-2y-1=0$ is

If $e$ and $e'$ are the eccentricities of a hyperbola and its conjugate hyperbola respectively,then $\frac{1}{e^2} + \frac{1}{e'^2} = \dots$