Chapter 9.3, Problem 79E

To determine

To eliminate: Rotate the axes to eliminate the xy- term in the equation, them write the equation in standard form and sketch the graph of the resulting equation, showing both the sets of axes.

Answer to Problem 79E

The equation without xy term in standard form is ${\left(y\text{'}\right)}^2=4\left(x\text{'}-1\right)$ .

Explanation of Solution

Given information:

The given angle is $16x^2-24xy+9y^2-60x-80y+100=0$ .

Calculation:

The general second-degree equation

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

where $B\ne 0$ , can be written as

  $A\text{'}{\left(x\text{'}\right)}^2+C\text{'}{\left(y\text{'}\right)}^2+D\text{'}x\text{'}+E\text{'}y\text{'}+F\text{'}=0$

by rotating the coordinate axes through an angle $\theta$ , where

  $\cot 2\theta =\frac{A-C}{B}$

The coefficients of the new equation are obtained by making the substitution

  $x=x\text{'}\cos \theta -y\text{'}\sin \theta$ and $y=x\text{'}\sin \theta +y\text{'}\cos \theta$ .

Here, $A=16$ , $B=-24$ and $C=9$ . Therefore,

  $\begin{array}{l}\cot 2\theta =\frac{16-9}{-24}\\ \cot 2\theta =\frac{7}{-24}\\ \cot 2\theta =\frac{-7}{24}\\ \frac{{\cot }^2\theta -1}{2\cos \theta }=\frac{-7}{24}\\ 24\left({\cot }^2\theta -1\right)=-7\left(2\cos \theta \right)\\ 24{\cot }^2\theta -24=-14\cos \theta \\ 24{\cot }^2\theta -14\cos \theta -24=0\\ 2\left(3\cot \theta +4\right)\left(4\cot \theta -3\right)=0\\ \left(3\cot \theta +4\right)=0\text{ or }\left(4\cot \theta -3\right)=0\end{array}$

Consider

  $\begin{array}{l}3\cot \theta +4=0\\ 3\cot \theta =-4\\ \cot \theta =\frac{-4}{3}\end{array}$ or

  $\begin{array}{l}4\cot \theta -3=0\\ 4\cot \theta =3\\ \cot \theta =\frac{3}{4}\\ \tan \theta =\frac{4}{3}\end{array}$

Since, $\cot \theta =\frac{-4}{3}$ is not possible , therefore,

  $\begin{array}{l}\tan \theta =\frac{4}{3}\\ \theta ={\tan }^{-1}\frac{4}{3}\end{array}$ .

The equation in the $x\text{'}y\text{'}$ -coordinate system is obtained by making the substitutions:

  $\begin{array}{l}x=x\text{'}\cos \left({\tan }^{-1}\frac{4}{3}\right)-y\text{'}\sin \left({\tan }^{-1}\frac{4}{3}\right)\\ x=x\text{'}\times \frac{3}{5}-y\text{'}\times \frac{4}{5}\\ x=\frac{3x\text{'}-4y\text{'}}{5}\mathrm{......}\text{ (1)}\end{array}$ and

  $\begin{array}{l}y=x\text{'}\sin \left({\tan }^{-1}\frac{4}{3}\right)+y\text{'}\cos \left({\tan }^{-1}\frac{4}{3}\right)\\ y=x\text{'}\times \frac{4}{5}+y\text{'}\times \frac{3}{5}\\ y=\frac{4x\text{'}+3y\text{'}}{5}\mathrm{......}\text{ (2)}\end{array}$

Substitute the value of x and y in the given equation and solve:

The sketch of the graph of the resulting equation, showing both the axes is drawn below:

  

Hence,

The equation without xy term in standard form is ${\left(y\text{'}\right)}^2=4\left(x\text{'}-1\right)$ .