If the eccentricity of a hyperbola is $\frac{5}{3}$,then the eccentricity of its conjugate hyperbola is

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 $x^2 - 2y^2 = 1$ lying on the positive $X$-axis. Let $P(-1, 1)$ be a given point. Then the area of the triangle formed by the line $PS$ with the coordinate axes is (in sq. units)

The Cartesian form of the curve given by $x = \frac{a}{2} (t + \frac{1}{t})$ and $y = \frac{a}{2} (t - \frac{1}{t})$,where $t$ is a parameter,is:

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

The radius of the director circle of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is