Chapter 8.4, Problem 32E

Solution

To find: the axis of rotation formulas that will transform $2x^2+\sqrt{3}xy+3y^2-2y+5=0$ to an equation in $\left(u,v\right)$ coordinates.

The axis of rotation formulas are $x=\frac{u}{2}-\frac{\sqrt{3}v}{2}$ and $y=\frac{\sqrt{3}u}{2}+\frac{v}{2}$ .

Given information:

The given quadratic equation is $2x^2+\sqrt{3}xy+3y^2-2y+5=0$ .

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 $C$ .

  $\begin{array}{l}A=2\\ B=\sqrt{3}\\ C=3\end{array}$

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

  $\begin{array}{c}\cot 2\theta =\frac{2-3}{\sqrt{3}}\\ \cot 2\theta =\frac{-1}{\sqrt{3}}\\ \cot 2\theta =-\frac{1}{\sqrt{3}}\end{array}$

Take inverse cotangent function and use the property ${\cot }^{-1}\left(\cot x\right)=x$ and solve for $\theta$

  $\begin{array}{c}{\cot }^{-1}\left(\cot 2\theta \right)={\cot }^{-1}\left(-\frac{1}{\sqrt{3}}\right)\\ 2\theta =\frac{2\pi }{3}\\ \theta =\frac{\pi }{3}\end{array}$

Substitute $\frac{\pi }{3}$ for $\theta$ into the above formulas to the values the values $x$ and $y$ in $\left(u,v\right)$ -coordinates.

  $\begin{array}{c}x=u\cos \left(\frac{\pi }{3}\right)-v\sin \left(\frac{\pi }{3}\right)\\ =u\cdot \frac{1}{2}-v\cdot \frac{\sqrt{3}}{2}\\ =\frac{u}{2}-\frac{\sqrt{3}v}{2}\end{array}$

  $\begin{array}{c}y=u\sin \left(\frac{\pi }{3}\right)+v\cos \left(\frac{\pi }{3}\right)\\ =u\cdot \frac{\sqrt{3}}{2}+v\cdot \frac{1}{2}\\ =\frac{\sqrt{3}u}{2}+\frac{v}{2}\end{array}$

Therefore, the required axis of rotation formulas are $x=\frac{u}{2}-\frac{\sqrt{3}v}{2}$ and $y=\frac{\sqrt{3}u}{2}+\frac{v}{2}$ .