Answer
See graph
Work Step by Step
We are given the hyperbola:
$\dfrac{(y+2)^2}{4}-\dfrac{(x-1)^2}{16}=1$
The equation is in the standard form:
$\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1$
The transverse axis is parallel to the $y$-axis.
Determine $h,k,a,b,c$:
$h=1$
$k=-2$
$a^2=4\Rightarrow a=\sqrt 4=2$
$b^2=16\Rightarrow b=\sqrt {16}=4$
$c^2=a^2+b^2$
$c^2=4+16$
$c^2=20$
$c=\sqrt{20}=2\sqrt 5$
Th centre of the hyperbola is:
$(h,k)=(1,-2)$
Determine the coordinates of the vertices:
$(h,k-a)=(1,-2-2)=(1,-4)$
$(h+a,k)=(1,-2+2)=(1,0)$
Determine the coordinates of the foci:
$(h,k-c)=(1,-2-2\sqrt 5)$
$(h,k+c)=(1,-2+2\sqrt 5)$
Determine the asymptotes:
$y-k=\pm\dfrac{a}{b}(x-h)$
$y+2=\pm\dfrac{2}{4}(x-1)$
$y+2=\pm\dfrac{1}{2}(x-1)$
Graph the hyperbola:
