Answer
See graph
Work Step by Step
We are given the hyperbola:
$x^2-y^2-2x-4y-4=0$
Bring the equation to the standard form:
$(x^2-2x+1)-(y^2+4y+4)-1+4-4=0$
$(x-1)^2-(y+2)^2=1$
The standard form of the equation is:
$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$
The transverse axis is parallel to the $x$-axis.
Determine $h,k,a,b,c$:
$h=1$
$k=-2$
$a^2=1\Rightarrow a=\sqrt 1=1$
$b^2=1\Rightarrow b=\sqrt 1=1$
$c^2=a^2+b^2$
$c^2=1+1$
$c^2=2$
$c=\sqrt{2}$
The centre of the hyperbola is:
$(h,k)=(1,-2)$
Determine the coordinates of the vertices:
$(h-a,k)=(1-1,-2)=(0,-2)$
$(h+a,k)=(1+1,-2)=(2,-2)$
Determine the coordinates of the foci:
$(h-c,k)=(1-\sqrt{2},-2)$
$(h+c,k)=(1+\sqrt{2},-2)$
Determine the asymptotes:
$y-k=\pm\dfrac{b}{a}(x-h)$
$y+2=\pm\dfrac{1}{1}(x-1)$
$y+2=\pm(x-1)$
Graph the hyperbola:
