Answer
See graph
Work Step by Step
We are given the hyperbola:
$9y^2-4x^2-18y+24x-63=0$
Bring the equation to the standard form:
$(9y^2-18y+9)-(4x^2-24x+36)-9+36-63=0$
$9(y^2-2y+1)-4(x^2-6x+9)-36=0$
$9(y-1)^2-4(x-3)^2=36$
$\dfrac{9(y-1)^2}{36}-\dfrac{4(x-3)^2}{36}=1$
$\dfrac{(y-1)^2}{4}-\dfrac{(x-3)^2}{9}=1$
The standard form of the equation is:
$\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1$
The transverse axis is parallel to the $y$-axis.
Determine $h,k,a,b,c$:
$h=3$
$k=1$
$a^2=4\Rightarrow a=\sqrt {4}=2$
$b^2=9\Rightarrow b=\sqrt {9}=3$
$c^2=a^2+b^2$
$c^2=4+9$
$c^2=13$
$c=\sqrt{13}$
Determine the centre:
$(h,k)=(3,1)$
Determine the coordinates of the vertices:
$(h,k-a)=(3,1-2)=(3,-1)$
$(h,k+a)=(3,1+2)=(3,3)$
Determine the coordinates of the foci:
$(h,k-c)=\left(3,1-\sqrt{13}\right)$
$(h,k+c)=\left(3,1+\sqrt{13}\right)$
Determine the asymptotes:
$y-k=\pm\dfrac{a}{b}(x-h)$
$y-1=\pm\dfrac{2}{3}(x-3)$
Graph the hyperbola:
