Chapter 8.4, Problem 26E

Solution

To find: the axis of rotation formulas that will transform $-5xy+22=0$ to an equation in $\left(u,v\right)$ coordinates.

The axis of rotation formulas are $x=\frac{u}{\sqrt{2}}-\frac{v}{\sqrt{2}}$ and $y=\frac{u}{\sqrt{2}}+\frac{v}{\sqrt{2}}$ .

Given information:

The given quadratic equation is $-5xy+22=0$ .

Formula used:

  $\cot 2\theta =\frac{A-C}{B}$ , where $0<\theta <\frac{\pi }{2}$ for the quadratic equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ .

Axis-rotation formulas are $x=u\cos \theta -v\sin \theta$ and $y=u\sin \theta +v\cos \theta$ .

Calculation:

Compare the given equation with quadratic equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ find the values of constants, $A,B,$ and $C$ .

  $\begin{array}{l}A=0\\ B=-5\\ C=0\end{array}$

Substitute the obtained values of $A,B,$ and $C$ into the formula, $\cot 2\theta =\frac{A-C}{B}$ and solve for angle $\theta$ .

  $\begin{array}{c}\cot 2\theta =\frac{0-0}{-5}\\ \cot 2\theta =0\\ 2\theta ={\cot }^{-1}\left(0\right)\\ 2\theta =\frac{\pi }{2}\end{array}$

Divide the obtained equation by 2 to obtain the value of $\theta$ .

  $\begin{array}{c}\frac{2\theta }{2}=\frac{\frac{\pi }{2}}{2}\\ \theta =\frac{\pi }{4}\end{array}$

Substitute $\frac{\pi }{4}$ for $\theta$ into the above formulas to the values the values $x$ and $y$ in $\left(u,v\right)$ -coordinates.

  $\begin{array}{c}x=u\cos \left(\frac{\pi }{4}\right)-v\sin \left(\frac{\pi }{4}\right)\\ =u\cdot \frac{1}{\sqrt{2}}-v\cdot \frac{1}{\sqrt{2}}\\ =\frac{u}{\sqrt{2}}-\frac{v}{\sqrt{2}}\end{array}$

  $\begin{array}{c}y=u\sin \left(\frac{\pi }{4}\right)+v\cos \left(\frac{\pi }{4}\right)\\ =u\cdot \frac{1}{\sqrt{2}}+v\cdot \frac{1}{\sqrt{2}}\\ =\frac{u}{\sqrt{2}}+\frac{v}{\sqrt{2}}\end{array}$

Therefore, the required axis of rotation formulas are $x=\frac{u}{\sqrt{2}}-\frac{v}{\sqrt{2}}$ and $x=\frac{u}{\sqrt{2}}+\frac{v}{\sqrt{2}}$ .