Answer
See graph
Work Step by Step
We are given the hyperbola:
$9x^2-4y^2=36$
Bring the equation to the standard form:
$\dfrac{9x^2}{36}-\dfrac{4y^2}{36}=1$
$\dfrac{x^2}{4}-\dfrac{y^2}{9}=1$
The transverse axis is parallel to the $x$-axis.
Determine $h,k,a,b,c$:
$h=0$
$k=0$
$a^2=4\Rightarrow a=\sqrt 4=2$
$b^2=9\Rightarrow b=\sqrt 9=3$
$c^2=a^2+b^2$
$c^2=4+9$
$c^2=13$
$c=\sqrt{13}$
Determine the coordinates of the vertices:
$(h-a,k)=(0-2,0)=(-2,0)$
$(h+a,k)=(0+2,0)=(2,0)$
Determine the coordinates of the co-vertices:
$(h,k-b)=(0,0-3)=(0,-3)$
$(h,k+b)=(0,0+3)=(0,3)$
Determine the coordinates of the foci:
$(h-c,k)=(0-\sqrt{13},0)=(-\sqrt{13},0)$
$(h+c,k)=(0+\sqrt{13},0)=(\sqrt{13},0)$
Graph the hyperbola:
