Question

Given the equation $41x^2-24xy+34y^2-25=0$ rotate the axes so that there is no $xy$-term. Analyze and graph the new equation.

Short answer

The equation represent an ellipse with a rotation of$0.99°$

Step-by-step solution

Step 1. Given Information

The given equation is $41x^2-24xy+34y^2-25=0$

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

Step 1. Finding acute angle

Here $A=41,B=-24,C=34$

$\cot \left(2\theta \right)=\frac{41-34}{-24}=-\frac{7}{24}$

where role="math" localid="1650435128455" $0°<2\theta <180°$

$$\begin{gathered} 2\theta =co{t}^{-1}\left(-\frac{7}{24}\right) \\ 2\theta =1.85 \\ \theta =0.925 \end{gathered}$$

Step 3. Finding x and y

$$\begin{gathered} x=x\text{'}\cos \theta -y\text{'}\sin \theta \\ x=x\text{'}\cos \left(0.925\right)-y\text{'}\sin \left(0.925\right) \\ x=0.9x\text{'}-0.9y\text{'} \end{gathered}$$

and

$$\begin{gathered} y=x\text{'}\sin \left(0.925\right)+y\text{'}\cos \left(0.925\right) \\ y=0.9x\text{'}+0.9y\text{'} \end{gathered}$$

Step 4. Substituting these values into the original equation 

$$\begin{gathered} 41x^2-24xy+34y^2-25=0 \\ 41{\left(0.9x\text{'}-0.9y\text{'}\right)}^2-24\left(0.9x\text{'}-0.9y\text{'}\right)\left(0.9x\text{'}+0.9y\text{'}\right)+34{\left(0.9x\text{'}+0.9y\text{'}\right)}^2-25=0 \\ 41\left(0.81x{\text{'}}^2-1.62x\text{'}y\text{'}+0.81y{\text{'}}^2\right)-24\left(0.81x{\text{'}}^2-0.81y{\text{'}}^2\right)+34\left(0.81x{\text{'}}^2+1.62x\text{'}y\text{'}+0.81y{\text{'}}^2\right)-25=0 \\ 33.21x{\text{'}}^2-66.42x\text{'}y\text{'}+33.21y{\text{'}}^2-19.44x{\text{'}}^2+19.44y{\text{'}}^2+27.54x{\text{'}}^2+55.08x\text{'}y\text{'}+27.54y{\text{'}}^2-25=0 \\ 41.31x{\text{'}}^2-11.34x\text{'}y\text{'}+80.19y{\text{'}}^2-25=0 \end{gathered}$$

Step 6. Graphing equation