Question

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

role="math" localid="1647182014777" $13x^2-6\sqrt{3}xy+7y^2-16=0$

Short answer

$\begin{array}{rcl}x& =& \frac{1}{2}\left(x^{\text{'}}-\sqrt{3}{y}^{\text{'}}\right)\\ y& =& \frac{1}{2}\left(\sqrt{3}{x}^2+y^{\text{'}}\right)\end{array}$

Step-by-step solution

Step 1. Given information

An equation is$13x^2-6\sqrt{3}xy+7y^2-16=0$.

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 $A=4,B=-4,C=1$in this equation to get width="11">θ

$\begin{array}{rcl}\cot \left(2\theta \right)& =& \frac{13-7}{-6\sqrt{3}}\\ & =& -\frac{1}{\sqrt{3}}\\ \cos \theta & =& \cos 60°=\frac{1}{2}\\ \sin \theta & =& \sin 60°=\frac{\sqrt{3}}{2}\\ x& =& x^{\text{'}}·\frac{1}{2}-y^{\text{'}}·\frac{\sqrt{3}}{2}\\ & =& \frac{1}{2}\left(x^{\text{'}}-\sqrt{3}{y}^{\text{'}}\right)\\ y& =& x^{\text{'}}÷\frac{\sqrt{3}}{2}+y^{\text{'}}·\frac{1}{2}\\ & =& \frac{1}{2}\left(\sqrt{3}{x}^2+y^{\text{'}}\right)\end{array}$