The slope of the tangent drawn from the point $(1,1)$ to the hyperbola $2x^2-y^2=4$ is

If $5x + 9 = 0$ is the directrix of the hyperbola $16x^2 - 9y^2 = 144,$ then its corresponding focus 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 $k=$

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

Let $S$ be the focus of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ lying on the positive $X$-axis and $P(5, y_1)$ be a point on the hyperbola. Then $SP =$

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