Chapter 9, Problem 53CR

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 53CR

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

Explanation of Solution

Given information:

The equation is $5x^2-2xy+5y^2+12=0$

Calculation:

Consider the given equation

Here, $A=5,B=-2\text{ and }C=5$

  $\cot 2\theta =\frac{A-C}{B}=\frac{5-5}{-2}=0$

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 $5x^2-2xy+5y^2+12=0$

  $\begin{array}{l}5x^2-2xy+5y^2+12=0\left[\text{Original equation}\right]\\ 5{\left[\frac{\sqrt{2}}{2}\left(x\text{'}-y\text{'}\right)\right]}^2-2\left[\frac{\sqrt{2}}{2}\left(x\text{'}-y\text{'}\right)\right]\left[\frac{\sqrt{2}}{2}\left(x\text{'}+y\text{'}\right)\right]+5{\left[\frac{\sqrt{2}}{2}\left(x\text{'}+y\text{'}\right)\right]}^2=-12\left[\text{Subtitute for }x\text{ and }y\right]\\ 5\left[\frac{1}{2}{\left(x\text{'}\right)}^2+\frac{1}{2}{\left(y\text{'}\right)}^2-x\text{'}y\text{'}\right]-{\left(x\text{'}\right)}^2+{\left(y\text{'}\right)}^2+5\left[\frac{1}{2}{\left(x\text{'}\right)}^2+x\text{'}y\text{'}+\frac{1}{2}{\left(y\text{'}\right)}^2\right]=-12\left[\text{Multiply}\right]\\ \text{ 4}{\left(x\text{'}\right)}^2+6{\left(y\text{'}\right)}^2=12\\ \frac{{\left(x\text{'}\right)}^2}{3}-\frac{{\left(y\text{'}\right)}^2}{2}=1\left[\text{Standard form}\right]\end{array}$

In the $x\text{'}y\text{'}$ system, this is the circle centered at the origin. To find the coordinates of the vertices in the $xy$ system, substitute the coordinates 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