If the equation $x+y+n=0$ represents a normal to the hyperbola $\frac{x^2}{6}-\frac{y^2}{2}=1$,then $n=$

The eccentricity of the hyperbola whose length of the latus rectum is equal to $8$ and the length of its conjugate axis is equal to half of the distance between its foci is:

Let $P(3,3)$ be a point on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. If the normal to it at $P$ intersects the $x$-axis at $(9,0)$ and $e$ is its eccentricity,then the ordered pair $(a^{2}, e^{2})$ is equal to

If the length of the transverse and conjugate axes of a hyperbola are $8$ and $6$ respectively,then the difference of the focal distances of any point on the hyperbola will be

If the latus rectum of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ subtends an angle of $60^{\circ}$ at the other focus,then the eccentricity of the hyperbola is

Let the foci of the hyperbola coincide with the foci of the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$. If the eccentricity of the hyperbola is $5$, then the length of its latus rectum is: