Answer
See graph
Work Step by Step
We are given the hyperbola:
$16y^2-9x^2=144$
Bring the equation to the standard form:
$\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1$
$\dfrac{16y^2}{144}-\dfrac{9x^2}{144}=1$
$\dfrac{y^2}{9}-\dfrac{x^2}{16}=1$
The transverse axis is parallel to the $y$-axis.
Determine $h,k,a,b,c$:
$h=0$
$k=0$
$a^2=9\Rightarrow a=\sqrt {9}=3$
$b^2=16\Rightarrow b=\sqrt {16}=4$
$c^2=a^2+b^2$
$c^2=9+16$
$c^2=25$
$c=\sqrt{25}$
$c=5$
Determine the coordinates of the vertices:
$(h,k-a)=(0,0-3)=(0,-3)$
$(h,k+a)=(0,0+3)=(0,3)$
Determine the coordinates of the co-vertices:
$(h-b,k)=(0-4,0)=(-4,0)$
$(h+b,k)=(0+4,0)=(4,0)$
Determine the coordinates of the foci:
$(h,k-c)=(0,0-5)=(0,-5)$
$(h,k+c)=(0,0+5)=(0,5)$
Graph the hyperbola: