Exemplo 4 de Algebra and Trigonometry, 12.4 Rotation of Axes

Graphing an Equation That Has No x′y′ Terms

Problema e solução em inglês, como no livro original.

Graph the following equation relative to the $x^{'} y^{'}$ system:

x^{2} + 12 x y - 4 y^{2} = 30

Resolva no papel primeiro. Depois abra a solução um passo de cada vez e pare assim que conseguir terminar sozinho.

Solução
First, we find $\cot (2 θ) .$

$x^{2} + 12 x y - 4 y^{2} = 20 \Rightarrow A = 1, B=12,andC=−4$

$\begin{array}{l} \cot (2 θ) = \frac{A - C}{B} \\ \cot (2 θ) = \frac{1 - (−4)}{12} \\ \cot (2 θ) = \frac{5}{12} \end{array}$

Because $\cot (2 θ) = \frac{5}{12},$ we can draw a reference triangle as in Figure 9.

Figure 9

$\cot (2 θ) = \frac{5}{12} = \frac{adjacent}{opposite}$

Thus, the hypotenuse is

$\begin{array}{r} 5^{2} + 12^{2} = h^{2} \\ 25 + 144 = h^{2} \\ 169 = h^{2} \\ h = 13 \end{array}$
Next, we find $\sin θ$ and $\cos θ.$ We will use half-angle identities.

$\begin{array}{l} \begin{array}{l} \sin θ=\sqrt{\frac{1 - \cos (2 θ)}{2}}=\sqrt{\frac{1 - \frac{5}{13}}{2}}=\sqrt{\frac{\frac{13}{13} - \frac{5}{13}}{2}}=\sqrt{\frac{8}{13} \cdot \frac{1}{2}}=\frac{2}{\sqrt{13}} \end{array} \\ \cos θ=\sqrt{\frac{1 + \cos (2 θ)}{2}}=\sqrt{\frac{1 + \frac{5}{13}}{2}}=\sqrt{\frac{\frac{13}{13} + \frac{5}{13}}{2}}=\sqrt{\frac{18}{13} \cdot \frac{1}{2}}=\frac{3}{\sqrt{13}} \end{array}$
Now we find $x$ and $y .$

$\begin{array}{l} x = x^{'} \cos θ-y^{'}\sin \\ x = x^{'} (\frac{3}{\sqrt{13}}) - y^{'} (\frac{2}{\sqrt{13}}) \\ x = \frac{3 x^{'} - 2 y^{'}}{\sqrt{13}} \end{array}$

and

$\begin{array}{l} y = x^{'} \sin θ+y^{'}\cos \\ y = x^{'} (\frac{2}{\sqrt{13}}) + y^{'} (\frac{3}{\sqrt{13}}) \\ y = \frac{2 x^{'} + 3 y^{'}}{\sqrt{13}} \end{array}$

Now we substitute $x = \frac{3 x^{'} - 2 y^{'}}{\sqrt{13}}$ and $y = \frac{2 x^{'} + 3 y^{'}}{\sqrt{13}}$ into $x^{2} + 12 x y - 4 y^{2} = 30.$

$\begin{array}{l} {(\frac{3 x^{'} - 2 y^{'}}{\sqrt{13}})}^{2} + 12 (\frac{3 x^{'} - 2 y^{'}}{\sqrt{13}}) (\frac{2 x^{'} + 3 y^{'}}{\sqrt{13}}) - 4 {(\frac{2 x^{'} + 3 y^{'}}{\sqrt{13}})}^{2} = 30 \\ (\frac{1}{13}) [{(3 x^{'} - 2 y^{'})}^{2} + 12 (3 x^{'} - 2 y^{'}) (2 x^{'} + 3 y^{'}) - 4 {(2 x^{'} + 3 y^{'})}^{2}] = 30 & Factor . \\ (\frac{1}{13}) [9 x^{'}^{2} - 12 x^{'} y^{'} + 4 y^{'}^{2} + 12 (6 x^{'}^{2} + 5 x^{'} y^{'} - 6 y^{'}^{2}) - 4 (4 x^{'}^{2} + 12 x^{'} y^{'} + 9 y^{'}^{2})] = 30 & Multiply . \\ (\frac{1}{13}) [9 x^{'}^{2} - 12 x^{'} y^{'} + 4 y^{'}^{2} + 72 x^{'}^{2} + 60 x^{'} y^{'} - 72 y^{'}^{2} - 16 x^{'}^{2} - 48 x^{'} y^{'} - 36 y^{'}^{2}] = 30 & Distribute . \\ (\frac{1}{13})[65 x^{'} 2 - 104 y^{'} 2]=30 & Combine like terms . \\ 65 x^{'}^{2} - 104 y^{'}^{2} = 390 & Multiply . \\ \frac{x^{'}^{2}}{6} - \frac{4 y^{'}^{2}}{15} = 1 & Divide by 390 . \end{array}$

Figure 10 shows the graph of the hyperbola $\frac{{x^{'}}^{2}}{6} - \frac{4 {y^{'}}^{2}}{15} = 1.$

Figure 10

Como foi?

Solução

Solução