Chapter 8.4, Problem 33E

Solution

To find: the axis of rotation formulas that will transform $24xy-7y^2-4x+5=0$ to an equation in $\left(u,v\right)$ coordinates.

The axis of rotation formulas are $x=\frac{4u}{5}-\frac{3v}{5}$ and $y=\frac{3u}{5}+\frac{4v}{5}$ .

Given information:

The given quadratic equation is $24xy-7y^2-4x+5=0$ .

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 $C$ .

  $\begin{array}{l}A=0\\ B=24\\ C=-7\end{array}$

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

  $\begin{array}{c}\cot 2\theta =\frac{0-\left(-7\right)}{24}\\ \cot 2\theta =\frac{7}{24}\end{array}$

It is known that when $\cot 2\theta =\frac{7}{24}$ , then $\cos 2\theta =\frac{7}{25}$ .

Apply the trigonometric identity $\cos \theta =\sqrt{\frac{1+\cos 2\theta }{2}}$ and substitute $\cos 2\theta =\frac{7}{25}$ into the identity and simplify.

  $\begin{array}{c}\cos \theta =\sqrt{\frac{1+\frac{7}{25}}{2}}\\ =\sqrt{\frac{32}{50}}\\ =\sqrt{\frac{16}{25}}\\ =\frac{4}{5}\end{array}$

Similarly, apply the trigonometric identity $\sin \theta =\sqrt{\frac{1-\cos 2\theta }{2}}$ and substitute $\cos 2\theta =\frac{7}{25}$ into the identity and simplify.

  $\begin{array}{c}\sin \theta =\sqrt{\frac{1-\frac{7}{25}}{2}}\\ =\sqrt{\frac{18}{50}}\\ =\sqrt{\frac{9}{25}}\\ =\frac{3}{5}\end{array}$

Substitute $\sin \theta =\frac{3}{5}$ and $\cos \theta =\frac{4}{5}$ in above equation to find the equation in terms of $x$ .

  $\begin{array}{c}x=u\cdot \frac{4}{5}-v\cdot \frac{3}{5}\\ =\frac{4u}{5}-\frac{3v}{5}\end{array}$

Substitute $\sin \theta =\frac{3}{5}$ and $\cos \theta =\frac{4}{5}$ in above equation to find the equation in terms of $y$ .

  $\begin{array}{c}y=u\cdot \frac{3}{5}+v\cdot \frac{4}{5}\\ =\frac{3u}{5}+\frac{4v}{5}\end{array}$

Therefore, the required equations are $x=\frac{4u}{5}-\frac{3v}{5}$ and $y=\frac{3u}{5}+\frac{4v}{5}$ .