College Algebra, Chapter 8, Review Exercises, Section Review Exercises, Problem 38

Identify the type of curve which is represented by the equation $\displaystyle 2x^2 + 4 = 4x + y^2 $
Find the foci and vertices(if any), and sketch the graph

$
\begin{equation}
\begin{aligned}
y^2 - 2(x^2 -2x) &= 4 && \text{Factor and group terms}\\
\\
y^2 - 2(x^2-2x+1) &= 4 -2 && \text{Complete the square; Add } \left( \frac{-2}{2} \right)^2 =1 \text{ on the left and subtract 2 from the right}\\
\\
y^2 - 2(x - 1)^2 &= 2 && \text{Perfect square}\\
\\
\frac{y^2}{2} - (x - 1)^2 &= 1 && \text{Divide by 2}
\end{aligned}
\end{equation}
$


The equation is hyperbola that has the form $\displaystyle \frac{(y - k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$ with center at $(h,k)$ and vertical transverse axis.
The graph of the shifted hyperbola is obtained from the graph of $\displaystyle \frac{y^2}{2} - x^2 = 1$ by shifting it 1 unit to the right. This gives us a
$a^2 = 2$ and $b^2 = 1$, so $a = \sqrt{2}, b = 1$ and $c = \sqrt{a^2 + b^2} = \sqrt{2+1} = \sqrt{3}$. Then, by applying transformation

$
\begin{equation}
\begin{aligned}
\text{center } & (h,k) && \rightarrow && (1,0)\\
\\
\text{vertices } & (0,a)&& \rightarrow && (0,\sqrt{2}) && \rightarrow && (0+1,\sqrt{2}) && = && (1, \sqrt{2})\\
\\
& (0,-a)&& \rightarrow && (0,-\sqrt{2}) && \rightarrow && (0+1,-\sqrt{2}) && = && (1,-\sqrt{2})\\
\\
\text{foci } & (0,c)&& \rightarrow && (0,\sqrt{3}) && \rightarrow && (0+1,\sqrt{3}) && = && (1,\sqrt{3})\\
\\
& (0,-c)&& \rightarrow && (0,-\sqrt{3}) && \rightarrow && (0+1,-\sqrt{3}) && = && (1,-\sqrt{3})
\end{aligned}
\end{equation}
$

Then, the graph is