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.$34x^2-24xy+41y^2-25=0$

Short answer

The ellipse is rotated through an angle of$-36.{87}^{\circ }$.

$$\begin{gathered} x=0.8x\text{'}+0.6y\text{'} \\ y=-0.6x\text{'}+0.8y\text{'} \end{gathered}$$

Step-by-step solution

Step 1. Given Information

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

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

Step 2. Finding acute angle 

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

$$\begin{gathered} cot2\theta =\frac{34-41}{-24}=\frac{7}{24} \\ 2\theta =-73.74 \\ \theta =-36.{87}^{\circ } \end{gathered}$$

Step 3. Finding expression for x,y

$$\begin{gathered} x=x\text{'}\cos \left(-36.87\right)-y\text{'}\sin \left(-36.87\right) \\ x=0.79x\text{'}+0.6y\text{'} \\ and \\ y=x\text{'}\sin \left(-36.87\right)+y\text{'}\cos \left(-36.87\right) \\ y=-0.6x\text{'}+0.79y\text{'} \end{gathered}$$

Step 4. Substitute these values in the original Equation

$$\begin{gathered} 34x^2-24xy+41y^2-25=0 \\ 34{\left(0.79x\text{'}+0.6y\text{'}\right)}^2-24\left(0.79x\text{'}+0.6y\text{'}\right)\left(-0.6x\text{'}+0.9y\text{'}\right)+41{\left(-0.6x\text{'}+0.9y\text{'}\right)}^2-25=0 \\ 34\left(0.6x{\text{'}}^2+0.9x\text{'}y\text{'}+0.3y{\text{'}}^2\right)-24\left(-0.47x{\text{'}}^2+0.7x\text{'}y\text{'}-0.3x\text{'}y\text{'}+0.5y{\text{'}}^2\right)+41\left(0.8y{\text{'}}^2-1.08x\text{'}y\text{'}+0.3x{\text{'}}^2\right)-25=0 \end{gathered}$$

Step 5. Simplify the above Equation

$$\begin{gathered} 20.4x{\text{'}}^2+30.6x\text{'}y\text{'}+10.2y{\text{'}}^2+11.2x{\text{'}}^2-9.6x\text{'}y\text{'}-12y{\text{'}}^2+32.8y{\text{'}}^2-44.2x\text{'}y\text{'}+13.2x{\text{'}}^2-25=0 \\ 44.8x{\text{'}}^2-23.2x\text{'}y\text{'}+31y{\text{'}}^2=25 \end{gathered}$$

Step 6. Graphing Equation