Chapter A5.3, Problem 21E

To determine

To identify: the graph of the equation.

Answer to Problem 21E

The graph of the equation isparallel horizontal lines

  ${\left(y^{\prime }\right)}^2=1$

Explanation of Solution

Given:

  $x^2-2xy+y^2=2$

Calculation:

Given the following equation:

  $x^2-2xy+y^2=2$

Comparing with the standard equation for a Quadratic Curve,

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

Thus,

  $A=1,B=-2,C=1$

For the cross-product term to vanish, the coordinate axes must be rotated by an angle alpha, suchthat

  $\cot 2\alpha =\frac{A-C}{B}$

  $=\frac{1-1}{-2}$

  $=0$

Therefore,

  $2\alpha =\frac{\pi }{2}$

  $\Rightarrow \alpha =\frac{\pi }{4}$

After rotation of coordinate axes, the original coordinates, $x$ and y, are replaced by newcoordinates, $x^{\prime }$ and $y^{\prime }$ , such that

  $\begin{array}{l}x=x^{\prime }\cos \alpha -y^{\prime }\sin \alpha \\ =\frac{\sqrt{2}}{2}{x}^{\prime }-\frac{\sqrt{2}}{2}{y}^{\prime }\\ y=x^{\prime }\sin \alpha +y^{\prime }\cos \alpha \\ =\frac{\sqrt{2}}{2}{x}^{\prime }+\frac{\sqrt{2}}{2}{y}^{\prime }\end{array}$

Substituting the above values of $x$ and $y$ in the given Quadratic Equation,

  ${\left(\frac{\sqrt{2}}{2}{x}^{\prime }-\frac{\sqrt{2}}{2}{y}^{\prime }\right)}^2-2\left(\frac{\sqrt{2}}{2}{x}^{\prime }-\frac{\sqrt{2}}{2}{y}^{\prime }\right)\left(\frac{\sqrt{2}}{2}{x}^{\prime }+\frac{\sqrt{2}}{2}{y}^{\prime }\right)+{\left(\frac{\sqrt{2}}{2}{x}^{\prime }+\frac{\sqrt{2}}{2}{y}^{\prime }\right)}^2=2$

  $\Rightarrow \left(\frac{1}{2}{\left(x^{\prime }\right)}^2-x^{\prime }{y}^{\prime }+\frac{1}{2}{\left(y^{\prime }\right)}^2\right)-2\left(\frac{1}{2}{\left(x^{\prime }\right)}^2-\frac{1}{2}{x}^{\prime }{y}^{\prime }+\frac{1}{2}{x}^{\prime }{y}^{\prime }-\frac{1}{2}{\left(y^{\prime }\right)}^2\right)+\left(\frac{1}{2}{\left(x^{\prime }\right)}^2+x^{\prime }{y}^{\prime }+\frac{1}{2}{\left(y^{\prime }\right)}^2\right)=2$

  $\Rightarrow 2{\left(y^{\prime }\right)}^2=2$

  $\Rightarrow {\left(y^{\prime }\right)}^2=1$

Which are parallel horizontal lines.

Conclusion:

The graph of the equation isparallel horizontal lines

  ${\left(y^{\prime }\right)}^2=1$