College Algebra, Chapter 8, 8.4, Section 8.4, Problem 32

Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find the center, foci, vertices and lengths of the major and minor axes. If it is a parabola, find the vertex, focus and directrix. If it is a hyperbola, find the center, foci, vertices and asymptotes. Sketch the graph of the equation. If the equation has no graph, explain why.


$
\begin{equation}
\begin{aligned}

x^2 - y^2 =& 10(x - y) + 1
&& \text{Distribute } 10
\\
\\
x^2 - y^2 =& 10x - 10y + 1
&& \text{Group terms}
\\
\\
(x^2 - 10x + \quad ) - (y^2 - 10y + \quad ) =& 1
&& \text{Complete the square: add } \left( \frac{-10}{2} \right)^2 = 25 \text{ and subtract $25$ on the right side}
\\
\\
(x^2 - 10x + 25) - (y^2 - 10y + 25) =& 1 + 25 - 25
&& \text{Perfect square}
\\
\\
(x - 5)^2 - (y - 5)^2 =& 1
&&

\end{aligned}
\end{equation}
$


The equation is a shifted hyperbola with center at $(5,5)$ and horizontal transverse axis. It is derived from the hyperbola $x^2 - y^2 = 1$. Since $a^2 = 1$ and $b^2 = 1$, we have $a = 1, b = 1$ and $c = \sqrt{1 + 1} = \sqrt{2}$. Thus, the foci lie $\sqrt{2}$ units to the right and left of the center. Consequently, the vertices of the hyperbola lies $1$ unit to the left and right of the center. By applying transformations, we get

foci

$\displaystyle (5, 5) \to (5 + \sqrt{2}, 5)$

$\displaystyle (5, 5) \to (5 - \sqrt{2}, 5)$

vertices

$\displaystyle (5, 5) \to (5 + 1, 5) = (6, 5)$

$\displaystyle (5, 5) \to (5 - 1, 5) = (4, 5)$

Also, the asymptotes of the unshifted hyperbola are $\displaystyle y = \pm \frac{b}{a} x = \pm x$, so, the asymptotes of the shifted hyperbola are


$
\begin{equation}
\begin{aligned}

y - 5 =& \pm (x - 5)
\\
\\
y - 5 =& \pm x \mp 5
\\
\\
y =& x \text{ and } y = -x + 10

\end{aligned}
\end{equation}
$


Therefore, the graph is