The tangent drawn at an extremity (in the first quadrant) of 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. If $O$ is the origin, then $(O A)^2-(O B)^2=$

If the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ passing through the point $(4,6)$ is 2 , then the equation of the tangent to this hyperbola at $(4,6)$ is

A hyperbola passes through the point $P(\sqrt{2}, \sqrt{3})$ and has foci at $( \pm 2,0)$. Then, the point that lies on the tangent drawn to this hyperbola at $P$ is

Let $P(a \sec \theta, b \tan \theta)$ and $Q(a \sec \phi, b \tan \phi)$, where $\theta+\phi=\frac{\pi}{2}$ be two points on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ If $(h, k)$ is the point of intersection of the normals drawn at $P$ and $Q$ then $K=$