Chapter 9, Problem 6CT

To determine

To rotate: the axes to eliminate the xy -term of $xy-x+3y+1=0$ . Then write the equation in standard form and which conic do the equation represents.

Answer to Problem 6CT

The equation in standard form is $\frac{X^2}{2}-\frac{Y^2}{2}+\sqrt{2}X+2\sqrt{2}Y+1=0$ in the form of standard hyperbola.

Explanation of Solution

Given information: Given equation is $xy-x+3y+1=0$ .

Calculation:

Compare the given equation with general equation:

  $\begin{array}{c}A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0\\ A=0\\ B=1\\ C=0\\ \text{Find}\text{the}\text{value}\text{of}\text{θ}\text{as}\text{follows:}\\ \text{cot2}\theta \text{=}\frac{A-C}{B}=\frac{0-0}{1}\\ 2\theta =\frac{\pi }{2}\\ \theta =\frac{\pi }{4}\end{array}$

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

  $\begin{array}{l}x=X\cos \frac{\pi }{4}-Y\sin \frac{\pi }{4}\\ x=\frac{X}{\sqrt{2}}-\frac{Y}{\sqrt{2}}=\frac{X-Y}{\sqrt{2}}\end{array}$

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

  $\begin{array}{l}y=X\sin \frac{\pi }{4}+Y\cos \frac{\pi }{4}\\ y=\frac{X}{\sqrt{2}}+\frac{Y}{\sqrt{2}}=\frac{X+Y}{\sqrt{2}}\end{array}$

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

  $\begin{array}{l}\left(\frac{X-Y}{\sqrt{2}}\right)\left(\frac{X+Y}{\sqrt{2}}\right)-\left(\frac{X-Y}{\sqrt{2}}\right)+3\left(\frac{X+Y}{\sqrt{2}}\right)+1=0\\ \frac{X^2}{2}-\frac{Y^2}{2}+\sqrt{2}X+2\sqrt{2}Y+1=0\end{array}$

Therefore, the equation in standard form is $\frac{X^2}{2}-\frac{Y^2}{2}+\sqrt{2}X+2\sqrt{2}Y+1=0$ in the form of standard hyperbola.