Question

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

$11x^2+10\sqrt{3}xy+y^2-4=0$

Short answer

$$\begin{gathered} x=\frac{\sqrt{3}{x}^{\text{'}}}{2}-\frac{y^{\text{'}}}{2} \\ y=\frac{x^{\text{'}}}{2}+\frac{\sqrt{3}{y}^{\text{'}}}{2} \end{gathered}$$

Step-by-step solution

Step 1. Given information

An equation is$11x^2+10\sqrt{3}xy+y^2-4=0$.

Step 2. Determine the appropriate rotation formulas.

The given conic is given by the equation given below.

$11x^2+10\sqrt{3}xy+y^2-4=0$

The rotation formulae for the given conic such that $xy-$term is absent are given by the formulae given below, where $\theta$can be determined by the relation $\cot 2\theta =\frac{A-C}{B}$

$$\begin{gathered} x=x^{\text{'}}\cos \theta -y^{\text{'}}\sin \theta \\ y=x^{\text{'}}\sin \theta +y^{\text{'}}\cos \theta \end{gathered}$$

The value of the $\theta$can be determined by substituting the values of $A,B,C$in the expression that evaluates the value of angle $\theta$

$\begin{array}{rcl}\cot 2\theta & =& \frac{11-1}{10\sqrt{3}}\\ & =& \frac{\pi }{6}\end{array}$

Therefore the rotation formulae for the given equation of conic are given by the equations that follow.

$\begin{array}{rcl}x& =& x^{\text{'}}\cos \frac{\pi }{6}-y^{\text{'}}\sin \frac{\pi }{6}\\ & =& \frac{\sqrt{3}{x}^{\text{'}}}{2}-\frac{y^{\text{'}}}{2}\\ y& =& x^{\text{'}}\sin \frac{\pi }{6}+y^{\text{'}}\cos \frac{\pi }{6}\\ & =& \frac{x^{\text{'}}}{2}+\frac{\sqrt{3}{y}^{\text{'}}}{2}\end{array}$