Question

Which of the following is INCORRECT for the hyperbola $x^2-2y^2-2x+8y-1=0$

• A. Its eccentricity is $\sqrt{2}$
• B. Length of the transverse axis is $2\sqrt{3}$
• C. Length of the conjugate axis is $2\sqrt{6}$
• D. Latus rectum is $4\sqrt{3}$

Answer 1

Answer: (A)

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 $\frac{(y-2{)}^2}{3}-\frac{(x-1{)}^2}{6}=1\dots (1)$
or $\frac{Y^2}{3}-\frac{X^2}{6}=1$
where $X=x-1$ and $Y=y-2\dots (2)$
$\therefore$ 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 $2a$, 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 $2b$,
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 $=\frac{2b^2}{a}=\frac{2\times 6}{\sqrt{3}}=4\sqrt{3}$.