Question

In Problems 31– 42, rotate the axes so that the new equation contains no $xy$-term. Analyze and graph the new equation. Refer to Problems 21–30 for Problems 31– 40.

$5x^2+6xy+5y^2-8=0$

Short answer

$$\begin{gathered} x=\frac{1}{\sqrt{2}}x\text{'}-\frac{1}{\sqrt{2}}y\text{'} \\ y=\frac{1}{\sqrt{2}}x\text{'}+\frac{1}{\sqrt{2}}y\text{'} \end{gathered}$$

Step-by-step solution

Step 1. Given Information

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

We need to determine the appropriate rotation formula so that new equation does not contain $xy$term.

Step 2. Finding acute angle

Here $A=5,B=6,C=5$

$cot2\theta =\frac{5-5}{6}=0$

Where

$$\begin{gathered} cot2\theta =0 \\ 2\theta =90 \\ \theta ={45}^{\circ } \end{gathered}$$

Step 3. Finding the values of x and y

$$\begin{gathered} x=x\text{'}\cos {45}^{\circ }-y\text{'}\sin {45}^{\circ } \\ x=\frac{1}{\sqrt{2}}x\text{'}-\frac{1}{\sqrt{2}}y\text{'} \end{gathered}$$

and

$$\begin{gathered} y=x\text{'}\sin {45}^{\circ }+y\text{'}\cos {45}^{\circ } \\ y=\frac{1}{\sqrt{2}}x\text{'}+\frac{1}{\sqrt{2}}y\text{'} \end{gathered}$$

Step 4. Substituting these values into the original equation 

$5x^2+6xy+5y^2-8=0$

$$\begin{gathered} \frac{5}{2}{\left(x\text{'}-y\text{'}\right)}^2+6\times \frac{1}{\sqrt{2}}\left(x\text{'}-y\text{'}\right)\times \frac{1}{\sqrt{2}}\left(x\text{'}+y\text{'}\right)+\frac{5}{2}{\left(x\text{'}-y\text{'}\right)}^2-8=0 \\ \frac{5}{2}\left(x{\text{'}}^2-2x\text{'}y\text{'}+y{\text{'}}^2\right)+3\left(x{\text{'}}^2-y{\text{'}}^2\right)+\frac{5}{2}\left(x{\text{'}}^2+2x\text{'}y\text{'}+y{\text{'}}^2\right)=8 \end{gathered}$$

Step 5. Simplify the above Equation

$$\begin{gathered} \frac{5}{2}\left(x{\text{'}}^2-2x\text{'}y\text{'}+y{\text{'}}^2\right)+3\left(x{\text{'}}^2-y{\text{'}}^2\right)+\frac{5}{2}\left(x{\text{'}}^2+2x\text{'}y\text{'}+y{\text{'}}^2\right)=8 \\ \frac{5}{2}x{\text{'}}^2-5x\text{'}y\text{'}+\frac{5}{2}y{\text{'}}^2+3x{\text{'}}^2-3y{\text{'}}^2+\frac{5}{2}x{\text{'}}^2+5x\text{'}y\text{'}+\frac{5}{2}y{\text{'}}^2=8 \\ 5x{\text{'}}^2+5y{\text{'}}^2+3x{\text{'}}^2-3y{\text{'}}^2=8 \\ 8x{\text{'}}^2+2y{\text{'}}^2=8 \\ x{\text{'}}^2+\frac{1}{4}y{\text{'}}^2=1 \\ \frac{x{\text{'}}^2}{1}+\frac{y{\text{'}}^2}{4}=1 \end{gathered}$$

This represent equation of ellipse.

Step 6. Graphing Equation