The equation of the hyperbola is
$x^{2}-2 y^{2}-2 x+8 y-1=0$
or $ (x-1)^{2}-2(y-2)^{2}+6=0$
or $ \frac{(x-1)^{2}}{-6}+\frac{(y-2)^{2}}{3}=1$
or $ \frac{(y-2)^{2}}{3}-\frac{(x-1)^{2}}{6}=1 \ldots (1)$
or $ \frac{Y^{2}}{3}-\frac{X^{2}}{6}=1$
where $X = x -1$ and $Y = y -2 \ldots(2)$
$\therefore $ The centre $=(0,0)$ in the $X - Y$ co-ordinates.
$\therefore $ The centre $=(1,2)$ in the $x-y$ co-ordinates, using $(2)$.
If the transverse axis be of length $2 a$, then $a =\sqrt{3}$,
since in the equation (1) the
transverse axis is parallel to the $y$-axis.
If the conjugate axis is of length $2 b$,
then $b =\sqrt{6}$.
But $b^{2}=a^{2}\left(e^{2}-1\right)$
$\therefore 6=3\left( e ^{2}-1\right), $
$\therefore e ^{2}=3$
or $e =\sqrt{3}$
The length of the transverse axis $=2 \sqrt{3}$.
The length of the conjugate axis $=2 \sqrt{6}$.
Latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 6}{\sqrt{3}}=4 \sqrt{3}$.