Answer
$\dfrac{(y-1)^2}{9}-\dfrac{(x+2)^2}{16}=1$
Vertical transverse axis
Work Step by Step
We are given the hyperbola:
Center: $(-2,1)$
Focus: $(-2,6)$
Vertex: $(-2,4)$
As the $x$-coordinate is the same for center, focus and vertex, the standard equation of the hyperbola is:
$\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1$
From the center, we determine $h,k$:
$(h,k)=(-2,1)$
$h=-2$
$k=1$
Determine $a$ using the vertex:
$(-2,4)=(h,k+a)$
$k+a=4$
$1+a=4$
$a=3$
Determine $c$ using the focus:
$(-2,6)=(h,k+c)$
$k+c=6$
$1+c=6$
$c=5$
Determine $b$:
$c^2=a^2+b^2$
$b^2=c^2-a^2$
$b^2=5^2-3^2$
$b^2=16$
$b=\sqrt {16}=4$
The equation of the hyperbola is:
$\dfrac{(y-1)^2}{3^2}-\dfrac{(x+2)^2}{4^2}=1$
$\dfrac{(y-1)^2}{9}-\dfrac{(x+2)^2}{16}=1$
The traverse axis is vertical.
