Chapter 9.3, Problem 79E

To determine

To rotate: The axes to eliminate the $xy$ -term in the equation, write the equation in standard form and sketch the graph of the resulting equation.

Answer to Problem 79E

The standard form of the equation is $x\text{'}=-{\left(y\text{'}\right)}^2$ .

Explanation of Solution

Given information:

The equation is $3x^2-2\sqrt{3}xy+y^2+2x+2\sqrt{3}y=0$ .

Calculation:

Compare the equation with general equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ .

  $\begin{array}{c}A=3\\ B=-2\sqrt{3}\\ C=1\end{array}$

Find the value of $\theta$ as follows.

  $\begin{array}{c}\cot 2\theta =\frac{A-C}{B}\\ =\frac{3-2}{-2\sqrt{3}}\\ \cot 2\theta =-\frac{1}{\sqrt{3}}\\ \frac{{\left(\cot \theta \right)}^2-1}{2\cot \theta }=-\frac{1}{\sqrt{3}}\end{array}$

Simplify further.

  $\begin{array}{c}\sqrt{3}{\left(\cot \theta \right)}^2+2\cot \theta -\sqrt{3}=0\\ -\frac{1}{\sqrt{3}}\left(\sqrt{3}-3\cot \theta \right)\left(\cot \theta +\sqrt{3}\right)=0\\ \cot \theta =\frac{\sqrt{3}}{3}\\ \theta =\frac{\pi }{3}\end{array}$

Substitute $\frac{\pi }{3}$ for $\theta$ in the equation $x=x\text{'}\cos \theta -y\text{'}\sin \theta$ to find $x$ coordinate.

  $\begin{array}{c}x=x\text{'}\cos \left(\frac{\pi }{3}\right)-y\text{'}\sin \left(\frac{\pi }{3}\right)\\ =\frac{x\text{'}}{2}-\frac{\sqrt{3}y\text{'}}{2}\\ =\frac{x\text{'}-\sqrt{3}y\text{'}}{2}\end{array}$

Substitute $\frac{\pi }{3}$ for $\theta$ in the equation $y=x\text{'}\sin \theta +y\text{'}\cos \theta$ to find $y$ coordinate.

  $\begin{array}{c}y=x\text{'}\sin \left(\frac{\pi }{3}\right)+y\text{'}\cos \left(\frac{\pi }{3}\right)\\ =\frac{\sqrt{3}x\text{'}}{2}+\frac{y\text{'}}{2}\\ =\frac{\sqrt{3}x\text{'}+y\text{'}}{2}\end{array}$

Substitute the equations for x and y into the given equation to eliminate the xy term.

  $\begin{array}{c}\left[\begin{array}{l}3{\left(\frac{x\text{'}-\sqrt{3}y\text{'}}{2}\right)}^2-2\sqrt{3}\left(\frac{x\text{'}-\sqrt{3}y\text{'}}{2}\right)\left(\frac{\sqrt{3}x\text{'}+y\text{'}}{2}\right)+\\ {\left(\frac{\sqrt{3}x\text{'}+y\text{'}}{2}\right)}^2+2\left(\frac{x\text{'}-\sqrt{3}y\text{'}}{2}\right)+2\sqrt{3}\left(\frac{\sqrt{3}x\text{'}+y\text{'}}{2}\right)\end{array}\right]=0\\ x\text{'}+{\left(y\text{'}\right)}^2=0\\ x\text{'}=-{\left(y\text{'}\right)}^2\end{array}$

Sketch the graph of the equation as follows.

  

Therefore, the standard form of the equation is $x\text{'}=-{\left(y\text{'}\right)}^2$ .