If $\frac{x^2}{\alpha+3}+\frac{y^2}{2-\alpha}=1$ represents a hyperbola,then $\alpha$ lies in

Find the equation of the tangent to the hyperbola $\frac{x^2}{3} - \frac{y^2}{2} = 1$ which is equally inclined to the axes.

The magnitude of the gradient of the tangent at an extremity of the latera recta of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is equal to (where $e$ is the eccentricity of the hyperbola).

$PQ$ is a double ordinate of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ such that $\triangle OPQ$ is an equilateral triangle, where $O$ is the centre of the hyperbola. Then the eccentricity $e$ of the hyperbola satisfies:

For the hyperbola $x^2 \sec^2 \theta - y^2 \csc^2 \theta = 1$,which of the following remains constant when $\theta$ varies?

The equation of the hyperbola whose asymptotes are the lines $3x+4y-2=0$ and $2x+y+1=0$ and which passes through the point $(1,1)$ is