Find the length of the latus rectum of the hyperbola $16x^2 - 9y^2 = 144$. (in $/3$)

Let $x+y+1=0$ and $x-y+4=0$ be the asymptotes of a hyperbola $H$. If $(1,1)$ is a point on $H$,then the length of the latus rectum of $H$ is

The foci of a hyperbola are $(\pm 2, 0)$ and its eccentricity is $\frac{3}{2}$. $A$ tangent,perpendicular to the line $2x + 3y = 6$,is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the $x$- and $y$-axes are $a$ and $b$ respectively,then $|6a| + |5b|$ is equal to $..........$.

If the normal drawn to the hyperbola $xy=16$ at $(8,2)$ meets the hyperbola again at a point $(\alpha, \beta)$,then $|\beta|+\frac{1}{|\alpha|}=$

Let the eccentricity $e$ of a hyperbola satisfy the equation $6e^2 - 11e + 3 = 0$. If the foci of the hyperbola are $(3, 5)$ and $(3, -4)$, then the length of its latus rectum is:

The equation of a line passing through the centre of a rectangular hyperbola is $x - y - 1 = 0$. If one of the asymptotes is $3x - 4y - 6 = 0$,the equation of the other asymptote is: