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

The equation of the common tangents to the two hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are-

Let $m_1$ and $m_2$ be the slopes of the tangents drawn from the point $P (4,1)$ to the hyperbola $H: \frac{y^2}{25}-\frac{x^2}{16}=1$. If $Q$ is the point from which the tangents drawn to $H$ have slopes $\left| m _1\right|$ and $\left| m _2\right|$ and they make positive intercepts $\alpha$ and $\beta$ on the $x$ axis, then $\frac{(P Q)^2}{\alpha \beta}$ is equal to $............$.

If a hyperbola passes through the point $\mathrm{P}(10,16)$ and it has vertices at $(\pm 6,0),$ then the equation of the normal to it at $P$ is

The eccentricity of the conic ${x^2} - 4{y^2} = 1$, is

Centre of hyperbola $9{x^2} - 16{y^2} + 18x + 32y - 151 = 0$ is