Find the center, transverse axis, vertices, foci, and asymptotes. Graph each equation by hand. Verify your graph using a graphing utility.

$y^2-x^2-4y+4x-1=0$

The given function is

$y^2-x^2-4y+4x-1=0$

The hyperbola with center $\left(h,k\right)$and transverse axis parallel to $y$-axis is given by

$\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$; $b^2=c^2-a^2$

where foci are at $\left(h,k\pm c\right)$, vertices are at $\left(h,k\pm a\right)$and asymptotes are

$y-k=\pm \frac{a}{b}\left(x-h\right)$

The given function is

$y^2-x^2-4y+4x-1=0$

Convert into standard form of the hyperbolas.

Grouping terms,

localid="1647360257385" $\left(y^2-4y\right)-\left(x^2-4x\right)=1$

Complete squaring,

$\left(y^2-4y+4\right)-\left(x^2-4x+4\right)=1$

${\left(y-2\right)}^2-{\left(x-2\right)}^2=1$

This the standard form of hyperbola.

So, $a=1,b=1$

$h=2,k=2$

Since $c^2=a^2+b^2$,

$c^2=1+1$

$c^2=2$

$c=\sqrt{2}$

Thus, the center of the hyperbola is$\left(h,k\right)\equiv \left(2,2\right)$

The foci are $\left(h,k\pm c\right)\equiv \left(2,2\pm \sqrt{2}\right)$,

The vertices are $\left(h,k+a\right)\equiv \left(2,2+1\right)\equiv \left(2,3\right)$and

$\left(h,k-a\right)\equiv \left(2,2-1\right)\equiv \left(2,1\right)$

Since foci and vertices all lie on the line $x=2$, the transverse axis is parallel to $y$-axis.

The asymptotes are

$y-2=\pm \frac{1}{1}\left(x-2\right)$

So, $y-1=\pm \left(x-2\right)$

The graph of the hyperbola is