The line $3x - 4y = 5$ is a tangent to the hyperbola $x^2 - 4y^2 = 5$. The point of contact is

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

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 latus-rectum of the hyperbola $16x^2 - 9y^2 = 144$ is

$A$ hyperbola passes through the point $P(\sqrt{2}, \sqrt{3})$ and has foci at $(\pm 2, 0)$. Then the point that lies on the tangent drawn to this hyperbola at $P$ is

If the line $lx + my = 1$ is a normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,then $\frac{a^2}{l^2} - \frac{b^2}{m^2}$ is equal to