The length of the latus rectum of the hyperbola $9x^2 - 16y^2 - 18x - 32y - 151 = 0$ is

Let $P(x_0, y_0)$ be the point on the hyperbola $3x^2 - 4y^2 = 36$ which is nearest to the line $3x + 2y = 1$. Then $\sqrt{2}(y_0 - x_0)$ is equal to:

For the hyperbola $x^2-y^2-4x+2y+c=0$,if the focus is $S(2+2\sqrt{2}, k)$ and the directrix that is adjacent to $S$ is $x=2+\sqrt{2}$,then $c=$

The equation $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos \theta$ represents:

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 $k=$

Let $0 < \theta < \frac{\pi}{2}$. If the eccentricity of the hyperbola $\frac{x^2}{\cos^2 \theta} - \frac{y^2}{\sin^2 \theta} = 1$ is greater than $2$,then the length of its latus rectum lies in the interval