Exemplo 3 de Algebra and Trigonometry, 12.4 Rotation of Axes

Rewriting an Equation with respect to the x′ and y′ axes without the x′y′ Term

Problema e solução em inglês, como no livro original.

Rewrite the equation $8 x^{2} - 12 x y + 17 y^{2} = 20$ in the $x^{'} y^{'}$ system without an $x^{'} y^{'}$ term.

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

Solução
First, we find $\cot (2 θ) .$ See Figure 7.

$\begin{array}{l} 8 x^{2} - 12 x y + 17 y^{2} = 20 \Rightarrow A = 8, B = - 12 and C = 17 \\ \cot (2 θ) = \frac{A - C}{B} = \frac{8 - 17}{- 12} \\ \cot (2 θ) = \frac{- 9}{- 12} = \frac{3}{4} \end{array}$

Figure 7

$\cot (2 θ) = \frac{3}{4} = \frac{adjacent}{opposite}$

So the hypotenuse is

$\begin{array}{r} 3^{2} + 4^{2} = h^{2} \\ 9 + 16 = h^{2} \\ 25 = h^{2} \\ h = 5 \end{array}$

Next, we find $\sin θ$ and $\cos θ.$
$\begin{array}{l} \begin{array}{l} \sin θ=\sqrt{\frac{1 - \cos (2 θ)}{2}}=\sqrt{\frac{1 - \frac{3}{5}}{2}}=\sqrt{\frac{\frac{5}{5} - \frac{3}{5}}{2}}=\sqrt{\frac{5 - 3}{5} \cdot \frac{1}{2}}=\sqrt{\frac{2}{10}}=\sqrt{\frac{1}{5}} \end{array} \\ \sin θ=\frac{1}{\sqrt{5}} \\ \cos θ=\sqrt{\frac{1 + \cos (2 θ)}{2}}=\sqrt{\frac{1 + \frac{3}{5}}{2}}=\sqrt{\frac{\frac{5}{5} + \frac{3}{5}}{2}}=\sqrt{\frac{5 + 3}{5} \cdot \frac{1}{2}}=\sqrt{\frac{8}{10}}=\sqrt{\frac{4}{5}} \\ \cos θ=\frac{2}{\sqrt{5}} \end{array}$

Substitute the values of $\sin θ$ and $\cos θ$ into $x = x^{'} \cos θ-y^{'}\sin$ and $y = x^{'} \sin θ+y^{'}\cos$

$\begin{array}{l} \begin{array}{l} x = x^{'} \cos θ-y^{'}\sin \\ x = x^{'} (\frac{2}{\sqrt{5}}) - y^{'} (\frac{1}{\sqrt{5}}) \\ x = \frac{2 x^{'} - y^{'}}{\sqrt{5}} \end{array} \end{array}$

and

$\begin{array}{l} \begin{array}{l} y = x^{'} \sin θ+y^{'}\cos \end{array} \\ y = x^{'} (\frac{1}{\sqrt{5}}) + y^{'} (\frac{2}{\sqrt{5}}) \\ y = \frac{x^{'} + 2 y^{'}}{\sqrt{5}} \end{array}$

Substitute the expressions for $x$ and $y$ into in the given equation, and then simplify.

$\begin{array}{l} 8 {(\frac{2 x^{'} - y^{'}}{\sqrt{5}})}^{2} - 12 (\frac{2 x^{'} - y^{'}}{\sqrt{5}}) (\frac{x^{'} + 2 y^{'}}{\sqrt{5}}) + 17 {(\frac{x^{'} + 2 y^{'}}{\sqrt{5}})}^{2} = 20 \\ 8(\frac{(2 x^{'} - y^{'}) (2 x^{'} - y^{'})}{5})-12(\frac{(2 x^{'} - y^{'}) (x^{'} + 2 y^{'})}{5})+17(\frac{(x^{'} + 2 y^{'}) (x^{'} + 2 y^{'})}{5})=20 \\ 8 (4 x^{'}^{2} - 4 x^{'} y^{'} + y^{'}^{2}) - 12 (2 x^{'}^{2} + 3 x^{'} y^{'} - 2 y^{'}^{2}) + 17 (x^{'}^{2} + 4 x^{'} y^{'} + 4 y^{'}^{2}) = 100 \\ 32 x^{'}^{2} - 32 x^{'} y^{'} + 8 y^{'}^{2} - 24 x^{'}^{2} - 36 x^{'} y^{'} + 24 y^{'}^{2} + 17 x^{'}^{2} + 68 x^{'} y^{'} + 68 y^{'}^{2} = 100 \\ 25 x^{'}^{2} + 100 y^{'}^{2} = 100 \\ \frac{25}{100} x^{'}^{2} + \frac{100}{100} y^{'}^{2} = \frac{100}{100} \end{array}$
Write the equations with $x^{'}$ and $y^{'}$ in the standard form with respect to the new coordinate system.

$\frac{{x^{'}}^{2}}{4} + \frac{{y^{'}}^{2}}{1} = 1$

Figure 8 shows the graph of the ellipse.

Figure 8

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O Try It do livro logo depois deste exemplo: mesma ideia, outros números. Só a resposta é dada.

Rewrite the $13 x^{2} - 6 \sqrt{3} x y + 7 y^{2} = 16$ in the $x^{'} y^{'}$ system without the $x^{'} y^{'}$ term.

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