Chapter 9.3, Problem 80E

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 80E

The standard form of the equation is $\frac{{\left(x\text{'}\right)}^2}{5/8}+\frac{{\left(y\text{'}\right)}^2}{3/8}=1$ .

Explanation of Solution

Given information:

The equation is $2x^2+xy+2y^2-8=0$ .

Calculation:

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

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

Find the value of $\theta$ as follows.

  $\begin{array}{c}\cot 2\theta =\frac{2-2}{1}\\ =\frac{0}{1}\\ 2\theta =\frac{\pi }{2}\\ \theta =\frac{\pi }{4}\end{array}$

Substitute $\frac{\pi }{4}$ for $\theta$ in the equation $x=x\text{'}\cos \theta -y\text{'}\sin \theta$ to find $x$ coordinate.

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

Substitute $\frac{\pi }{4}$ for $\theta$ in the equation $y=x\text{'}\sin \theta +y\text{'}\cos \theta$ to find $y$ coordinate.

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

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

  $\begin{array}{c}2{\left(\frac{x\text{'}-y\text{'}}{\sqrt{2}}\right)}^2+\left(\frac{x\text{'}-y\text{'}}{\sqrt{2}}\right)2\left(\frac{x\text{'}+y\text{'}}{\sqrt{2}}\right)+2{\left(\frac{x\text{'}+y\text{'}}{\sqrt{2}}\right)}^2-8=0\\ 2\left[2{\left(x\text{'}\right)}^2+2{\left(y\text{'}\right)}^2\right]+{\left(x\text{'}\right)}^2-{\left(y\text{'}\right)}^2-8=0\\ 5{\left(x\text{'}\right)}^2+3{\left(y\text{'}\right)}^2=8\\ \frac{{\left(x\text{'}\right)}^2}{5/8}+\frac{{\left(y\text{'}\right)}^2}{3/8}=1\end{array}$

In the $x\text{'}y\text{'}$ -system, this is a hyperbola centered at the origin with vertices at $\left(\pm \sqrt{\frac{5}{8}},0\right)$ . To find the coordinates of the vertices in the $xy$ -system, substitute the coordinates $\left(\pm \sqrt{\frac{5}{8}},0\right)$ into the equations $\frac{x\text{'}-y\text{'}}{\sqrt{2}}$ and $\frac{x\text{'}+y\text{'}}{\sqrt{2}}$ .

This substitution yields the vertices $\left(\frac{\sqrt{5}}{2\sqrt{6}},\frac{\sqrt{5}}{2\sqrt{6}}\right)$ and $\left(-\frac{\sqrt{5}}{2\sqrt{6}},-\frac{\sqrt{5}}{2\sqrt{6}}\right)$ in the $xy$ -system and also the asymptote of the hyperbola have the equation $y\text{'}=\pm \frac{3}{5}x$ .

Sketch the graph of the equation as follows.

  

Therefore, the standard form of the equation is $\frac{{\left(x\text{'}\right)}^2}{5/8}+\frac{{\left(y\text{'}\right)}^2}{3/8}=1$ .