The length of the latus rectum of the hyperbola $\frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 4$ is (where $\alpha \neq \frac{n\pi}{2}, n \in I$).

The eccentricity of the hyperbola $x^2 - y^2 = 25$ is

Two points $A$ and $B$ have coordinates $(1, 0)$ and $(-1, 0)$ respectively,and $Q$ is a point which satisfies the relation $AQ - BQ = \pm 1$. The locus of $Q$ is:

$A$ hyperbola passes through the points $(3, 2)$ and $(-17, 12)$ and has its centre at the origin and its transverse axis along the $x$-axis. The length of its transverse axis is:

Let $LL^{\prime}$ be the latus rectum through the focus $S$ of a hyperbola and $A^{\prime}$ be the opposite vertex of the hyperbola. If triangle $A^{\prime}LL^{\prime}$ is equilateral,then the eccentricity of the hyperbola is

Consider the hyperbola $H : x^2-y^2=1$ and a circle $S$ with center $N(x_2, 0)$. Suppose that $H$ and $S$ touch each other at a point $P(x_1, y_1)$ with $x_1 > 1$ and $y_1 > 0$. The common tangent to $H$ and $S$ at $P$ intersects the $x$-axis at point $M$. If $(l, m)$ is the centroid of the triangle $\triangle PMN$,then the correct expression$(s)$ is(are):
$(A) \frac{dl}{dx_1} = 1 - \frac{1}{3x_1^2}$ for $x_1 > 1$
$(B) \frac{dm}{dx_1} = \frac{x_1}{3\sqrt{x_1^2-1}}$ for $x_1 > 1$
$(C) \frac{dl}{dx_1} = 1 + \frac{1}{3x_1^2}$ for $x_1 > 1$
$(D) \frac{dm}{dy_1} = \frac{1}{3}$ for $y_1 > 0$