The auxiliary circle of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by:

The equation of the hyperbola whose directrix is $x + 2y = 1$,focus $(2, 1)$ and eccentricity $e = 2$ is:

Let $P(a \sec \theta, b \tan \theta)$ and $Q(a \sec \phi, b \tan \phi)$ where $\theta + \phi = \frac{\pi}{2}$ be two points on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. If $(h, k)$ is the point of intersection of the normals drawn at $P$ and $Q$,then $k=$

If for a hyperbola the ratio of the length of the conjugate axis to the length of the transverse axis is $3:2$,then the ratio of the distance between the foci to the distance between the two directrices is

The tangent drawn at an extremity (in the first quadrant) of the latus rectum of the hyperbola $\frac{x^2}{4}-\frac{y^2}{5}=1$ meets the $x$-axis and $y$-axis at $A$ and $B$ respectively. If $O$ is the origin,then $(OA)^2-(OB)^2=$

If the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ has eccentricity $e = \frac{5}{4}$ and the length of the latus rectum equal to $9$,then $ab$ is equal to