If the equation of the tangent to the hyperbola $5x^2 - 9y^2 - 20x - 18y - 34 = 0$ which makes an angle of $45^{\circ}$ with the positive $X$-axis is $x + by + c = 0$,then $b^2 + c^2 =$

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

One of the latus recta of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ subtends an angle $2 \operatorname{Tan}^{-1}\left(\frac{3}{2}\right)$ at the centre of the hyperbola. If $b^2=36$ and $e$ is the eccentricity of the given hyperbola,then $\sqrt{a^2+e^2}=$

If the area of the quadrilateral formed by the tangents drawn at the ends of the latus rectum of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is equal to the square of the distance between the center and one focus of the hyperbola,then $e^3$ is ($e$ is the eccentricity of the hyperbola).

The equation of the tangent at the point $(a \sec \theta, b \tan \theta)$ of the conic $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is:

$A$ double ordinate $PQ$ of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is such that $\Delta OPQ$ is equilateral, where $O$ is the centre of the hyperbola. Then the eccentricity $e$ satisfies the relation: