Example 2 from Algebra and Trigonometry, 12.4 Rotation of Axes

Finding a New Representation of an Equation after Rotating through a Given Angle

Find a new representation of the equation $2 x^{2} - x y + 2 y^{2} - 30 = 0$ after rotating through an angle of $θ = 45° .$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
Find $x$ and $y,$ where $x = x^{'} \cos θ-y^{'}\sin$ and $y = x^{'} \sin θ+y^{'}\cos$

Because $θ = 45°,$

$\begin{array}{l} x = x^{'} \cos (45°) - y^{'} \sin (45°) \\ x = x^{'} (\frac{1}{\sqrt{2}}) - y^{'} (\frac{1}{\sqrt{2}}) \\ x = \frac{x^{'} - y^{'}}{\sqrt{2}} \end{array}$

and

$\begin{array}{l} \begin{array}{l} y = x^{'} \sin (45°) + y^{'} \cos (45°) \\ y = x^{'} (\frac{1}{\sqrt{2}}) + y^{'} (\frac{1}{\sqrt{2}}) \\ y = \frac{x^{'} + y^{'}}{\sqrt{2}} \end{array} \end{array}$

Substitute $x = x^{'} \cos θ - y^{'} \sin θ$ and $y = x^{'} \sin θ+y^{'}\cos$ into $2 x^{2} - x y + 2 y^{2} - 30 = 0.$

$2 {(\frac{x^{'} - y^{'}}{\sqrt{2}})}^{2} - (\frac{x^{'} - y^{'}}{\sqrt{2}}) (\frac{x^{'} + y^{'}}{\sqrt{2}}) + 2 {(\frac{x^{'} + y^{'}}{\sqrt{2}})}^{2} - 30 = 0$

Simplify.

$\begin{array}{l} 2 \frac{(x^{'} - y^{'}) (x^{'} - y^{'})}{2} - \frac{(x^{'} - y^{'}) (x^{'} + y^{'})}{2} + 2 \frac{(x^{'} + y^{'}) (x^{'} + y^{'})}{2} - 30 = 0 & \begin{matrix} \end{matrix} FOIL method \\ x^{'}^{2} {- 2 x^{'} y}^{'} + y^{'}^{2} - \frac{(x^{'}^{2} - y^{'}^{2})}{2} + x^{'}^{2} + 2 x^{'} y^{'} + y^{'}^{2} - 30 = 0 & \begin{matrix} \end{matrix} Combine like terms . \\ 2 x^{'}^{2} + 2 y^{'}^{2} - \frac{(x^{'}^{2} - y^{'}^{2})}{2} = 30 & \begin{matrix} \end{matrix} Combine like terms . \\ 2 (2 x^{'}^{2} + 2 y^{'}^{2} - \frac{(x^{'}^{2} - y^{'}^{2})}{2}) = 2 (30) & \begin{matrix} \end{matrix} Multiply both sides by 2 . \\ 4 x^{'}^{2} + 4 y^{'}^{2} - (x^{'}^{2} - y^{'}^{2}) = 60 & \begin{matrix} \end{matrix} Simplify . \\ 4 x^{'}^{2} + 4 y^{'}^{2} - x^{'}^{2} + y^{'}^{2} = 60 & \begin{matrix} \end{matrix} Distribute . \\ \frac{3 x^{'}^{2}}{60} + \frac{5 y^{'}^{2}}{60} = \frac{60}{60} & \begin{matrix} \end{matrix} Set equal to 1 . \end{array}$

Write the equations with $x^{'}$ and $y^{'}$ in the standard form.

$\frac{{x^{'}}^{2}}{20} + \frac{{y^{'}}^{2}}{12} = 1$

This equation is an ellipse. Figure 6 shows the graph.

Figure 6

How did it go?

Solution

Solution