If the latus rectum of an hyperbola be 8 and eccentricity be $3/\sqrt 5 $, then the equation of the hyperbola is

The graph of the conic $ x^2 - (y - 1)^2 = 1$  has one tangent line with positive slope that passes through the origin. the point of tangency being $(a, b). $ Then  Length of the latus rectum of the conic is

The equation of the tangent to the hyperbola $2{x^2} - 3{y^2} = 6$ which is parallel to the line $y = 3x + 4$, is

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

Let $0 < \theta  < \frac{\pi }{2}$. If the eccentricity of the hyperbola $\frac{{{x^2}}}{{{{\cos }^2}\,\theta }} - \frac{{{y^2}}}{{{{\sin }^2}\,\theta }} = 1$ is greater than $2$, then the length of its latus rectum lies in the interval

If the product of the perpendicular distances from any point on the hyperbola $\frac{{{x^2}}}{{{a^2}}}\,\, - \,\,\frac{{{y^2}}}{{{b^2}}}\,\,\, = \,1$ of eccentricity $e =\sqrt 3 \,$ from its asymptotes is equal to $6$, then the length of the transverse axis of the hyperbola is