Chapter 9.3, Problem 73E

To determine

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

Answer to Problem 73E

The center is $\left(0,0\right)$ , vertices is $\left(0,\pm 4\right)$ , foci is $\left(0,\pm 2\sqrt{5}\right)$ and asymptotes is $y=\pm 2x$ .

Explanation of Solution

Given information:

The equation is $xy+1=0$ .

Calculation:

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

  $\begin{array}{c}A=0\\ B=1\\ C=0\end{array}$

Find the value of $\theta$ as follows.

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

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.

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

In the $x\text{'}y\text{'}$ -system, this is a hyperbola centered at the origin with vertices at $\left(0,\pm \sqrt{2}\right)$ . To find the coordinates of the vertices in the $xy$ -system, substitute the coordinates $\left(0,\pm \sqrt{2}\right)$ 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 $\left(-1,1\right)$ and $\left(1,-1\right)$ in the $xy$ -system and also the asymptote of the hyperbola have the equation $y\text{'}=\pm x\text{'}$ .

Sketch the graph of the equation as follows.

  

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