Chapter 9, Problem 52CR

To determine

To find: The standard form of the equation. Rotate the axes to eliminate the $xy$ -term in the equation and also sketch the graph of the resultant equation.

Answer to Problem 52CR

The $xy$ -term is eliminated by the rotation of axes. The standard form of the equation is $\frac{{\left(x\text{'}\right)}^2}{9}-\frac{{\left(y\text{'}\right)}^2}{3}=1$ . The graph is represented.

Explanation of Solution

Given information:

The equation is $x^2-4xy+y^2+9=0$

Calculation:

Consider the given equation

Here, $A=1,B=-4\text{ and }C=1$

  $\cot 2\theta =\frac{A-C}{B}=\frac{1-1}{-4}=0$

So,

  $\begin{array}{l}\cot 2\theta =0\\ \Rightarrow 2\theta =\frac{\pi }{2}\\ \Rightarrow \theta =\frac{\pi }{4}\end{array}$

The equation in the $x\text{'}y\text{'}$ system is obtain by making the substitutions

  $\begin{array}{l}x=x\text{'}\cos \frac{\pi }{4}-y\text{'}\sin \frac{\pi }{4}\\ x=x\text{'}\left(\frac{1}{\sqrt{2}}\right)-y\text{'}\left(\frac{1}{\sqrt{2}}\right)\\ x=\frac{x\text{'}-y\text{'}}{\sqrt{2}}\\ x=\frac{\sqrt{2}}{2}\left(x\text{'}-y\text{'}\right)\end{array}$

And,

  $\begin{array}{l}y=x\text{'}\sin \frac{\pi }{4}+y\text{'}\cos \frac{\pi }{4}\\ y=x\text{'}\left(\frac{1}{\sqrt{2}}\right)+y\text{'}\left(\frac{1}{\sqrt{2}}\right)\\ y=\frac{x\text{'}+y\text{'}}{\sqrt{2}}\\ y=\frac{\sqrt{2}}{2}\left(x\text{'}+y\text{'}\right)\end{array}$

Into the equation $x^2-4xy+y^2+9=0$

  $\begin{array}{l}{x}^2-4xy+y^2+9=0\left[\text{Original equation}\right]\\ {\left[\frac{\sqrt{2}}{2}\left(x\text{'}-y\text{'}\right)\right]}^2-4\left[\frac{\sqrt{2}}{2}\left(x\text{'}-y\text{'}\right)\right]\left[\frac{\sqrt{2}}{2}\left(x\text{'}+y\text{'}\right)\right]+{\left[\frac{\sqrt{2}}{2}\left(x\text{'}+y\text{'}\right)\right]}^2=-9\left[\text{Subtitute for }x\text{ and }y\right]\\ \frac{1}{2}{\left(x\text{'}\right)}^2+\frac{1}{2}{\left(y\text{'}\right)}^2-x\text{'}y\text{'}-2\left\{{\left(x\text{'}\right)}^2-{\left(y\text{'}\right)}^2\right\}+\frac{1}{2}{\left(x\text{'}\right)}^2+x\text{'}y\text{'}+\frac{1}{2}{\left(y\text{'}\right)}^2=-9\left[\text{Multiply}\right]\\ -{\left(x\text{'}\right)}^2+3{\left(y\text{'}\right)}^2=-9\\ \frac{{\left(x\text{'}\right)}^2}{9}-\frac{{\left(y\text{'}\right)}^2}{3}=1\left[\text{Standard form}\right]\end{array}$

In the $x\text{'}y\text{'}$ system, this is the hyperbola centered at the origin with the vertices at $\left(9,-3\right)$ . To find the coordinates of the vertices in the $xy$ system, substitute the coordinates $\left(9,-3\right)$ into the equations

  $x=\frac{x\text{'}-y\text{'}}{\sqrt{2}}$ And $y=\frac{x\text{'}+y\text{'}}{\sqrt{2}}$

This substitution yields the vertices $\left(1,1\right)\text{ and }\left(-1,-1\right)$ in the $xy$ system. Note also that the asymptotes of the hyperbola have the equations $\left(y\text{'}=\pm x\text{'}\right)$ which corresponds to the original $x\text{ and }y\text{ axes}$

The graph of the equation can be represented as