Chapter A5.3, Problem 17E

To determine

To identify: the graph of the equation.

Answer to Problem 17E

The graph of the equation isa hyperbola

  $[{\left(x^{\prime }\right)}^2-{\left(y^{\prime }\right)}^2=4]$

Explanation of Solution

Given:

  $xy=2$

Calculation:

Given the following equation:

  $xy=2$

Comparing with the standard equation for a Quadratic Curve,

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

We get:

  $A=0,B=1,C=0$

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

That

  $\cot 2\alpha =\frac{A-C}{B}$

  $\begin{array}{l}=\frac{0-0}{1}\hfill \\ =0\hfill \end{array}$

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 new

coordinates, $x^{\prime }$ 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,

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

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

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

  $\Rightarrow [{\left(x^{\prime }\right)}^2-{\left(y^{\prime }\right)}^2=4]$

Which is a hyperbola.

Conclusion:

The graph of the equation isa hyperbola

  $[{\left(x^{\prime }\right)}^2-{\left(y^{\prime }\right)}^2=4]$