The length of the conjugate axis of a hyperbola is greater than the length of the transverse axis. Then, the eccentricity $e$ is

The equation $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos \theta$ represents:

The locus of the mid-point of the chord of the hyperbola $3x^2 - 2y^2 + 4x - 6y = 0$ which is parallel to $y = 2x$ is:

The point $P(-2 \sqrt{6}, \sqrt{3})$ lies on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ having eccentricity $e = \frac{\sqrt{5}}{2}$. If the tangent and normal at $P$ to the hyperbola intersect its conjugate axis at the points $Q$ and $R$ respectively,then $QR$ is equal to :

Let $a$ and $b$ respectively be the semi-transverse and semi-conjugate axes of a hyperbola whose eccentricity satisfies the equation $9e^2 - 18e + 5 = 0$. If $S(5, 0)$ is a focus and $5x = 9$ is the corresponding directrix of this hyperbola,then $a^2 - b^2$ is equal to

If $\sqrt{5} y - \sqrt{8} = 0$ is the equation of the directrix of a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} + 1 = 0$ and $\frac{\sqrt{5}}{2}$ is its eccentricity,then $\frac{1}{a} =$