Question

In Problem, determine the appropriate rotation formulas to use so that the new equation contains no $xy-$term.

role="math" localid="1647180393519" $5x^2+6xy+5y^2-8=0$

Short answer

$$\begin{gathered} x=\frac{\sqrt{2}}{2}\left(x^{\text{'}}-y^{\text{'}}\right) \\ y=\frac{\sqrt{2}}{2}\left(x^{\text{'}}+y^{\text{'}}\right) \end{gathered}$$

Step-by-step solution

Step 1. Given information

An equation is$5x^2+6xy+5y^2-8=0$.

Step 2. Determine the appropriate rotation formulas.

The appropriate acute angle $\theta$through which to rotate the axes so that the $xy-$term is removed is $\cot \left(2\theta \right)=\frac{A-C}{B}$.

Substitute $A=1,B=4,C=1$in this equation to get $\theta$.

$\begin{array}{rcl}\cot \left(2\theta \right)& =& \frac{5-5}{6}\\ & =& 0\\ \cos \theta & =& \frac{1}{\sqrt{2}}·\frac{\sqrt{2}}{\sqrt{2}}\\ & =& \frac{\sqrt{2}}{2}\\ \sin \theta & =& \frac{1}{\sqrt{2}}·\frac{\sqrt{2}}{\sqrt{2}}\\ & =& \frac{\sqrt{2}}{2}\end{array}$

$\begin{array}{rcl}x& =& x^{\text{'}}·\frac{\sqrt{2}}{2}-y^{\text{'}}·\frac{\sqrt{2}}{2}\\ & =& \frac{\sqrt{2}}{2}\left(x^{\text{'}}-y^{\text{'}}\right)\\ y& =& x^{\text{'}}·\frac{\sqrt{2}}{2}+y^{\text{'}}·\frac{\sqrt{2}}{2}\\ & =& \frac{\sqrt{2}}{2}\left(x^{\text{'}}+y^{\text{'}}\right)\end{array}$