Answer:
Option A
Explanation:
The equation of the hyperbola is
$x^{2}-2y^{2}-2x+8y-1=0$
or $ (x-1)^{2}-2(y-2)^{2}+6=0$
or $\frac{(x-1)^{2}}{-6}+\frac{(y-2)^{2}}{3}=1$
or $\frac{(y-2)^{2}}{3}-\frac{(x-1)^{2}}{6}=1$ .....(i)
or $\frac{Y^{2}}{3}-\frac{X^{2}}{6}=1$
where X =x-1 and Y= y-2 ....(ii)
$\therefore$ The centre =(0,0,) in X-Y co-ordinates
$\therefore$ The centre = (1,2) in the x-y co-ordinates using (ii)
If the transverse axis be length 2a, than a = $\sqrt{3}$, since in the equation (i) the tansverse axis is parallel to the y-axis
If the conjugate axis is of length 2b n, then b= $\sqrt{6}$
But $b^{2}=a^{2}(e^{2}-1)$
$\therefore$ $6=3(e^{2}-1), \therefore e^{2}=3$ or $ e=\sqrt{3}$
The length of the transverse axis =2$\sqrt{3}$
The length of the conjugate axis =2$\sqrt{6}$
Latus rectum = $\frac{2b^{2}}{a}=\frac{2\times6}{\sqrt{3}}=4\sqrt{3}$
