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

Let the foci of a hyperbola $H$ coincide with the foci of the ellipse $E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$ and the eccentricity of the hyperbola $H$ be the reciprocal of the eccentricity of the ellipse $E$. If the length of the transverse axis of $H$ is $\alpha$ and the length of its conjugate axis is $\beta$,then $3 \alpha^2+2 \beta^2$ is equal to :

Let $P(6,3)$ be a point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. If the normal at the point $P$ intersects the $x$-axis at $(9,0)$,then the eccentricity of the hyperbola is

If $(4, 0)$ and $(-4, 0)$ are the vertices and $(6, 0)$ and $(-6, 0)$ are the foci of a hyperbola,then its eccentricity is

Find the coordinates of the foci and the vertices,the eccentricity,and the length of the latus rectum of the hyperbola: $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$.

The distance between the foci of a hyperbola is $16$ and its eccentricity is $\sqrt{2}$. Its equation is