The equation of the normal to the hyperbola $\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1$ at $( - 4,\;0)$ is

Let $P (3\, sec\,\theta , 2\, tan\,\theta )$ and $Q\, (3\, sec\,\phi , 2\, tan\,\phi )$ where $\theta + \phi \, = \frac{\pi}{2}$ , be two distinct points on the hyperbola $\frac{{{x^2}}}{9} - \frac{{{y^2}}}{4} = 1$ . Then the ordinate of the point of intersection of the normals at $P$ and $Q$ is

The line $2 \mathrm{x}+\mathrm{y}=1$ is tangent to the hyperbola $\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$. If this line passes through the point of intersection of the nearest directrix and the $\mathrm{x}$-axis, then the eccentricity of the hyperbola is

Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ be $\frac{5}{4}$. If the equation of the normal at the point $\left(\frac{8}{\sqrt{5}}, \frac{12}{5}\right)$ on the hyperbola is $8 \sqrt{5} x +\beta y =\lambda$, then $\lambda-\beta$ is equal to

Eccentricity of conjugate hyperbola of $16x^2 - 9y^2 - 32x - 36y - 164 = 0$ will be-

Length of latusrectum of curve $xy = 7x + 5y$ is