Question

If the pair of straight lines $x^2-y^2+6x+4y+5=0$ represents transverse and conjugate axes of the hyperbola and centre of hyberbola is $(\alpha ,\beta )$, then value of $\beta -\alpha$ is

Answer 1

$x^2-y^2+6x+4y+5=0.....\left(1\right)$
$\because x^2-y^2=(x-y)(x+y)$
Let the seperate equation of line be $(x-y+l)(x+y+m)$
In order to find value's of $l$ and $m$, we have to equate the coefficients of $x$ and $y$
$x^2-xy+lx+xy-y^2+ly+mx-my+lm=0$
$x^2-y^2+(l+m)x+(l-m)y+lm=0....\left(2\right)$
Comparing $\left(1\right)$ & $\left(2\right)$, we get
$l+m=6$
$l-m=4$
$2l=10$
$\therefore l=5$ and $m=1$
So equations of axes are $x-y+5=0$ and $x+y+1=0$
Centre will be point of intersection of both the lines, so $x=-3$ and $y=2$
$\therefore$ Centre is $(-3,2)=(\alpha ,\beta )$
$\Rightarrow \beta -\alpha =5$