Find the equation of the tangent to the hyperbola $x^2 - 4y^2 = 36$ which is perpendicular to the line $x - y + 4 = 0$.

Let the sum of the focal distances of the point $P(4,3)$ on the hyperbola $H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be $8 \sqrt{\frac{5}{3}}$. If for $H$,the length of the latus rectum is $l$ and the product of the focal distances of the point $P$ is $m$,then $9l^2 + 6m$ is equal to :-

Let $\lambda x - 2y = \mu$ be a tangent to the hyperbola $a^{2}x^{2} - y^{2} = b^{2}$. Then $\left(\frac{\lambda}{a}\right)^{2} - \left(\frac{\mu}{b}\right)^{2}$ is equal to

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

If $(1,2)$ is the focus,$x+2y=0$ is the directrix,and $\sqrt{2}$ is the eccentricity of a hyperbola,then the equation of the hyperbola is

Let $S = \{(x,y) \in \mathbb{R}^2 : \frac{y^2}{1+r} - \frac{x^2}{1-r} = 1\}$,where $r \neq \pm 1$. Then $S$ represents