The equation of the normal at the point $(6, 4)$ on the hyperbola $\frac{x^2}{9} - \frac{y^2}{16} = 3$ is:

If the foci of a hyperbola are the same as that of the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$ and the eccentricity of the hyperbola is $\frac{15}{8}$ times the eccentricity of the ellipse,then the smaller focal distance of the point $\left(\sqrt{2}, \frac{14}{3} \sqrt{\frac{2}{5}}\right)$ on the hyperbola is equal to

$A$ tangent to a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ intercepts a length of unity from each of the coordinate axes. Then the point $(a, b)$ lies on the rectangular hyperbola:

If $A$ and $B$ are two points in a plane,such that $|PA - PB| = \text{constant}$,then the locus of $P$ is

If the center,vertex,and focus of a hyperbola are $(0, 0)$,$(4, 0)$,and $(6, 0)$ respectively,then the equation of the hyperbola is:

Let $P(a \sec \theta, b \tan \theta)$ and $Q(a \sec \phi, b \tan \phi)$ be two points such that $\theta+\phi=\frac{\pi}{2}$ 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 at $P$ and $Q$,then $k=$