If $e_1$ and $e_2$ are respectively the eccentricities of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and its conjugate hyperbola,then the line $\frac{x}{2 e_1}+\frac{y}{2 e_2}=1$ touches the circle having centre at the origin. Find its radius.

Let $S$ be the focus of the hyperbola $\frac{x^2}{3}-\frac{y^2}{5}=1$,on the positive $x$-axis. Let $C$ be the circle with its centre at $A(\sqrt{6}, \sqrt{5})$ and passing through the point $S$. If $O$ is the origin and $SAB$ is a diameter of $C$,then the square of the area of the triangle $OSB$ is equal to ....................

The locus of the feet of the perpendiculars drawn from either focus onto a variable tangent to the hyperbola $16y^2 - 9x^2 = 1$ is

If the eccentricity of the hyperbola $x^2 - y^2 \sec^2 \alpha = 5$ is $\sqrt{3}$ times the eccentricity of the ellipse $x^2 \sec^2 \alpha + y^2 = 25$,then a value of $\alpha$ is:

If $3x + 2\sqrt{2}y + k = 0$ is a normal to the hyperbola $4x^2 - 9y^2 - 36 = 0$ making positive intercepts on both the axes,then $k=$ (in $\sqrt{2}$)

The equation of the hyperbola with focus $(1, 2)$,eccentricity $e = \sqrt{3}$,and directrix $2x + y = 1$ is given by: