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

If a directrix of a hyperbola centered at the origin and passing through the point $(4, -2 \sqrt{3})$ is $\sqrt{5}x = 4$ and $e$ is its eccentricity,then $e^2 =$

If the latus rectum of a hyperbola through one focus subtends an angle of $60^{\circ}$ at the other focus,then its eccentricity is

$A$ mirror in the first quadrant is in the shape of a hyperbola whose equation is $xy=1$. $A$ light source in the second quadrant emits a beam of light that hits the mirror at the point $(2, 1/2)$. If the reflected ray is parallel to the $Y$-axis,the slope of the incident beam is

Let $O$ be the origin,and $P$ and $Q$ be two points on the rectangular hyperbola $xy = 12$ such that the midpoint of the line segment $PQ$ is $(\frac{1}{2}, -\frac{1}{2})$. Then the area of the triangle $OPQ$ equals:

$A$ line parallel to the straight line $2x - y = 0$ is tangent to the hyperbola $\frac{x^{2}}{4} - \frac{y^{2}}{2} = 1$ at the point $(x_{1}, y_{1})$. Then $x_{1}^{2} + 5y_{1}^{2}$ is equal to: