I am working on rotation of conic sections and I'm having trouble determining the angle of rotation from the coefficients of the general conic equation. I'm given $$11x^2-24xy+4y^2+20=0$$ From this equation I know that $$cot(2\theta)= \frac{A-C}B=\frac{11-4}{-24}=\frac{7}{-24}$$ But when I draw a triangle with angle $2\theta$ to find $cos(2\theta)$, I end up with $$cos(2\theta)=\frac{7}{25}$$ Using this value for $cos(2\theta)$ with the half-angle formulas results in an incorrect solution, which does not eliminate the original $xy$-term. However, if I rewrite $$cot(2\theta)=\frac{-7}{24}$$ and draw a new triangle with angle $2\theta$, I end up with $$cos(2\theta)=\frac{-7}{25}$$ which yields the solution given in the back of the book after substitution and simplification. I keep encountering problems of this sort and am having trouble understanding why changing the sign of $$cot(2\theta)=\frac{7}{-24}$$ is allowed. I'm sure I am missing something with respect to the trig functions. I would be grateful for any clarification.
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See this post I made on converting a conic to standard form: http://math.stackexchange.com/a/1127581/97045 two things to note: (1) There are at least 4 possible angles that can represent the degree to which an axially aligned conic is rotated, separated by a quarter rotation and (2) the $s$ and $c$ variables in the post represent the sine and cosine of that angle respectively, so there are multiple valid $s$ and $c$ angles, you can choose one arbitrarily – DanielV Jun 05 '15 at 03:23
3 Answers
The matrix $\begin{bmatrix}\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{bmatrix}$ rotates $\begin{bmatrix}x&y\end{bmatrix}$ clockwise.
Therefore, $$ \begin{align} &\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{bmatrix}\begin{bmatrix}a&0\\0&b\end{bmatrix}\begin{bmatrix}\cos(\theta)&\sin(\theta)\\-\sin(\theta)&\cos(\theta)\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\\[6pt] &=\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}\color{#C00000}{a\cos^2(\theta)+b\sin^2(\theta)}&\color{#00A000}{(a-b)\sin(\theta)\cos(\theta)}\\\color{#00A000}{(a-b)\sin(\theta)\cos(\theta)}&\color{#0000F0}{a\sin^2(\theta)+b\cos^2(\theta)}\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\\[6pt] &=\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}\color{#C00000}{11}&\color{#00A000}{-12}\\\color{#00A000}{-12}&\color{#0000F0}{4}\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} \end{align} $$ says that if we rotate $\begin{bmatrix}x&y\end{bmatrix}$ clockwise by $\theta$ we get an upright conic (one whose axes are aligned with the coordinate axes).
To compute $\theta$, note that $11-4=(a-b)\cos(2\theta)$ and $-24=(a-b)\sin(2\theta)$. Therefore, $$ \tan(2\theta)=-\frac{24}{7}=\tan(-1.287) $$ Thus, if we rotate the graph of $11x^2-24xy+4y^2+20=0$ counterclockwise $0.6435$ radians, it will be upright.
Note that knowing $\tan(2\theta)$ only tells what $\theta$ is mod $\pi/2$. This is fine, however, since rotating an upright conic by $\pi/2$ also gives an upright conic.
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It may be more helpful to resolve the orientation issue using the double-angle formula for tangent, $ \ \tan(2\theta) \ = \ \frac{2 \ \tan \theta}{1 \ - \ \tan^2 \theta} \ $ . Working with the formula for the rotation angle of the conics, you have $ \ \tan (2 \theta) \ = \ - \frac{24}{7} \ $ (here, we don't worry about whether the numerator or the denominator should be negative), we obtain
$$ \frac{2 \ \tan \theta}{1 \ - \ \tan^2 \theta} \ \ = \ \ - \frac{24}{7} $$
$$ \Rightarrow \ \ 2 \ \tan \theta \ \ = \ \ - \frac{24}{7} \ + \ \frac{24}{7} \ \tan^2 \theta \ \ \Rightarrow \ \ 12 \ \tan^2 \theta \ - \ 7 \tan \theta \ - \ 12 \ \ = \ \ 0 $$
$$ \ \tan \theta \ = \ \frac{4}{3} \ , \ - \frac{3}{4} \ \ . $$
The ambiguity is unavoidable, but note that this gives the slopes of two perpendicular lines, one of which is the line for the "focal axis" of the conic section, the other being the perpendicular axis to that. This gives just two choices for eliminating the $ \ "xy" \ $ term in the conic equation. For this conic, which proves to be a hyperbola, the focal axis is along $ \ y \ = \ \frac{4}{3} \ x \ $ .
EDIT: It actually doesn't matter which choice you make for the slope of the line and the associated cosine and sine values, if all you want to do is eliminate the "mixed" term. The "correct" choice will rotate the axes so that the $ \ x'- $ axis runs through the conic's focal axis [ $ \ x'^2 \ - \ 4y'^2 \ = \ 4 \ $ ], while the "other" choice puts the $ \ y'-$ axis there [ $ \ y'^2 \ - \ 4x'^2 \ = \ 4 \ $ ]. (This is a point that I can't recall seeing mentioned in at least introductory texts. And most instructors don't have the patience to show the calculation worked out both ways...) If one desires to extract the properties of the conic, the differing alignments are not important.
EDIT 2: (fixed an algebra error in the alternate conic equation) In view of DanielV's added comment, one can consider the four angles as rotating the conic so that the "positive" direction on its focal axis is pointed along each ray of the lines with slope $ \ \frac{4}{3} \ $ or $ \ -\frac{3}{4} \ $.
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So there is no way to reconcile the sign issues while using cosine? I am more curious as to how the book generated the correct solution, even though the value of $cot(2\theta)=\frac{A-C}{B}=\frac{7}{-24}$ determined from the conic equation suggests that the denominator is negative, which leads to the wrong value for cosine. I want to avoid the "pick the value that works" method of solving these problems. The book makes no mention of using the tangent ideas, so I'm assuming they used some method to determine the "correct" value for cosine. Thanks for the advice. – Orestes95 Jun 05 '15 at 07:37
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I think it is safe to say that there is no need to "reconcile" the sign issue. As robjohn remarks in his answer, the function $ \ \cot(2 \theta) \ $ has a period of $ \ \frac{\pi}{2} \ $ , so the cotangent equation for the rotation angle of the conic generally has four solutions in the interval $ \ 0 \ \le \ \theta \ < \ 2 \pi \ $ . Two of these given angles of rotation put the focal axis on the conic on the $ \ x'-$ axis, whihe the other two angles lay the focal axis on the $ \ y'- \ $ axis . The textbook "correct" answer is one such choice, but not the only acceptable one. – colormegone Jun 06 '15 at 03:27
If you draw a triangle with adjacent $-7$ and opposite $24$, then by pythagorean theorem, the hypotenuse is $25$. Therefore the cosine is $\frac{-7}{25}$.
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