If $2x - ky + 3 = 0$ and $3x - y + 1 = 0$ are conjugate lines with respect to the hyperbola $5x^2 - 6y^2 = 15$,then $k =$

$A$ hyperbola passes through the point $P(\sqrt{2}, \sqrt{3})$ and has foci at $(\pm 2, 0)$. Then the point that lies on the tangent drawn to this hyperbola at $P$ is

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 $|m_1|$ and $|m_2|$ and they make positive intercepts $\alpha$ and $\beta$ on the $x$-axis,then $\frac{(PQ)^2}{\alpha \beta}$ is equal to $............$.

The area of the quadrilateral formed by the foci of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ and its conjugate hyperbola is (in square units):

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

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 $b^2 + c^2 =$