The equation of the normal to the hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$ at $(-4, 0)$ is

If the line $2x + \sqrt{6}y = 2$ touches the hyperbola $x^2 - 2y^2 = 4$,then the coordinates of the point of contact are

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 $SP =$

Tangents are drawn to the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$,parallel to the straight line $2x-y=1$. The points of contact of the tangents on the hyperbola are:
$(A) \left(\frac{9}{2\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
$(B) \left(-\frac{9}{2\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$
$(C) (3\sqrt{3}, -2\sqrt{2})$
$(D) (-3\sqrt{3}, 2\sqrt{2})$

The eccentricity of the hyperbola $-\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is:

The equation of the directrices of the conic $x^2 + 2x - y^2 + 5 = 0$ are