Question

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

width="139">x2-4xy+y2-3=0

Short answer

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

Step-by-step solution

Step 1. Given information

An equation is$x^2-4xy+y^2-3=0$.

Step 2. Determine the appropriate rotation formulas.

The given conic is given by the equation given below.

$x^2-4xy+y^2-3=0$

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 $\theta$.

$\begin{array}{rcl}\cot 2\theta & =& \frac{1-1}{-4}\\ & =& 0\end{array}$

$2\theta =\frac{\pi }{4}$

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

width="155">x=x'cosπ4-y'sinπ4=x'2-y'2

$\begin{array}{rcl}y& =& x^{\text{'}}\sin \frac{\pi }{4}+y^{\text{'}}\cos \frac{\pi }{4}\\ & =& \frac{x^{\text{'}}}{\sqrt{2}}+\frac{y^{\text{'}}}{\sqrt{2}}\end{array}$