# How do you graph y² - x² - 2y + 4x - 4 =0?

Jul 11, 2016

A hyperbole

$h \left(x , y\right) = \frac{1}{2} \left\{x - 2 , y - 1\right\} \left(\begin{matrix}- 2 & 0 \\ 0 & 2\end{matrix}\right) \left\{x - 2 , y - 1\right\} - 1$

#### Explanation:

f(x,y) = y² - x² - 2y + 4x - 4 =0 is a conic

The kind of conic can be answered after analyzing its associated quadratic form.

${q}_{f} \left(x , y\right) = \frac{1}{2} \left\{x , y\right\} \left(\begin{matrix}{f}_{x x} & {f}_{x y} \\ {f}_{y x} & {f}_{y y}\end{matrix}\right) \left\{x , y\right\}$

where ${f}_{{x}_{1} , {x}_{2}} = \frac{{\partial}^{2} f}{\partial {x}_{1} \partial {x}_{2}}$

Here

$\left(\begin{matrix}{f}_{x x} & {f}_{x y} \\ {f}_{y x} & {f}_{y y}\end{matrix}\right) = \left(\begin{matrix}- 2 & 0 \\ 0 & 2\end{matrix}\right)$ with eigenvalues $\left\{- 2 , 2\right\}$ which characterize this conic as a Hyperbole.

The hyperbole canonic representation is

$h \left(x , y\right) = \frac{1}{2} \left\{x - {x}_{0} , y - {y}_{0}\right\} \left(\begin{matrix}- 2 & 0 \\ 0 & 2\end{matrix}\right) \left\{x - {x}_{0} , y - {y}_{0}\right\} + {c}_{0}$

Now, comparing

$f \left(x , y\right) = h \left(x , y\right)$ we obtain the conditions:

{ (x_0^2 - y_0^2-4-c_0 = 0), (2 y_0-2 = 0), (4 - 2 x_0 =0) :}

Solving for ${x}_{0} , {y}_{0} , {c}_{0}$ gives

$\left\{{x}_{0} = 2 , {y}_{0} = 1 , {c}_{0} = - 1\right\}$

so the hyperbole is

$h \left(x , y\right) = \frac{1}{2} \left\{x - 2 , y - 1\right\} \left(\begin{matrix}- 2 & 0 \\ 0 & 2\end{matrix}\right) \left\{x - 2 , y - 1\right\} - 1$

centered at $\left\{2 , 1\right\}$

${\left(y - 1\right)}^{2} - {\left(x - 2\right)}^{2} - 1 = 0 \to \left(y - x + 1\right) \left(y + x - 3\right) - 1 = 0$

so the assymptotes are

$y - x + 1 = 0$ and $y + x - 3 = 0$