The directrix of the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ is:

Let the foci and length of the latus rectum of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ be $(\pm 5, 0)$ and $\sqrt{50}$,respectively. Then,the square of the eccentricity of the hyperbola $\frac{x^2}{b^2}-\frac{y^2}{a^2b^2}=1$ equals

Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be $2a$ and $2b$,respectively,and one focus and the corresponding directrix of this hyperbola be $(-5, 0)$ and $5x + 9 = 0$,respectively. If the product of the focal distances of a point $(\alpha, 2\sqrt{5})$ on the hyperbola is $p$,then $4p$ is equal to . . . . . . .

The tangent at an extremity (in the first quadrant) of the latus rectum of the hyperbola $\frac{x^2}{4} - \frac{y^2}{5} = 1$ meets the $x$-axis and $y$-axis at $A$ and $B$ respectively. Then $(OA)^2 - (OB)^2$,where $O$ is the origin,equals

If a hyperbola passes through the foci of the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$ and its transverse and conjugate axes coincide with the major and minor axes of the ellipse respectively,and the product of their eccentricities is $1$,then:

The asymptotes of a hyperbola are parallel to $2x + 3y = 0$ and $3x + 2y = 0$. The equation of the hyperbola whose center is at $(1, 2)$ and which passes through $(5, 3)$ is: