Question

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

$x^2+4xy+4y^2+5\sqrt{5}y+5=0$

Short answer

$\begin{array}{rcl}x& =& -\frac{3x^{\text{'}}}{5}+\frac{4y^{\text{'}}}{5}\\ y& =& -\frac{4x^{\text{'}}}{5}-\frac{3y^{\text{'}}}{5}\end{array}$

Step-by-step solution

Step 1. Given information

An equation is$x^2+4xy+4y^2+5\sqrt{5}y+5=0$.

Step 2. Determine the appropriate rotation formulas.

The given conic is given by the equation given below.

$x^2+4xy+4y^2+5\sqrt{5}y+5=0$

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 $\theta$.

width="157">cot2θ=1-44=-342θ=cot-1-34=12cot-1-34

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

$\begin{array}{rcl}x& =& x^{\text{'}}\cos \left({\cot }^{-1}\left(\frac{-3}{4}\right)\right)-y^{\text{'}}\sin \left({\cot }^{-1}\left(\frac{-3}{4}\right)\right)\\ & =& -\frac{3x^{\text{'}}}{5}+\frac{4y^{\text{'}}}{5}\\ y& =& x^{\text{'}}\sin \left({\cot }^{-1}\left(\frac{-3}{4}\right)\right)+y^{\text{'}}\cos \left({\cot }^{-1}\left(\frac{-3}{4}\right)\right)\\ & =& -\frac{4x^{\text{'}}}{5}-\frac{3y^{\text{'}}}{5}\end{array}$