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.

localid="1647365908255" $x^2+4xy+y^2-3=0$

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

For the given equation $x^2+4xy+y^2-3=0$. We need to determine the appropriate rotation so that the new equation does not contain $xy$term.

Step 1. Finding acute angle

Here $A=1,B=4,C=1$

$cot2\theta =\frac{1-1}{4}=0$

Where

$0^{\circ }<2\theta <{180}^{\circ }$

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

Step 3. Finding 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 

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

$$\begin{gathered} \frac{1}{2}{\left(x\text{'}-y\text{'}\right)}^2+4\left[\frac{1}{\sqrt{2}}\left(x\text{'}-y\text{'}\right)\right]\left[\frac{1}{\sqrt{2}}\left(x\text{'}+y\text{'}\right)\right]+\frac{1}{2}{\left(x\text{'}+y\text{'}\right)}^2-3=0 \\ \frac{1}{2}\left(x{\text{'}}^2-2x\text{'}y\text{'}+y{\text{'}}^2\right)+2\left(x{\text{'}}^2-y{\text{'}}^2\right)+\frac{1}{2}\left(x{\text{'}}^2+2x\text{'}y\text{'}+y{\text{'}}^2\right)=3 \end{gathered}$$

Step 5. Simplify above expression

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

The equation represent hyperbola.

Step 6. Graphing equation 

$\frac{x{\text{'}}^2}{1}-\frac{y{\text{'}}^2}{3}=1$