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.
[Math] Determining angle for rotation of conics
algebra-precalculusconic sectionstrigonometry
Related Solutions
It's not really clear what kind of relation OP expects, but here's something ...
Let the generators of a cone (with vertex at the origin and $z$-axis as axis) make angle $\alpha$ with the $xy$-plane. Let a cutting plane parallel to the $y$-axis make angle $\beta$ with the $xy$-plane, and let it meet the cone at $V := (-v \cos\alpha,0,v \sin\alpha)$ (which will be a vertex of the conic). That is, the cone and plane have equations
$$z^2 \cos^2\alpha = (x^2+y^2)\sin^2\alpha \qquad\qquad (x+v\cos\alpha)\sin\beta=(z-v\sin\alpha)\cos\beta$$
We can parameterize the intersection of the plane and cone by
$$P := p (\cos\alpha \cos\theta, \cos\alpha\sin\theta,\sin\alpha)$$
with $\theta$-dependent $p$ determined by substituting $P$ into the plane equation. The fully-parametric form of $P$ is then
$$P = \frac{v \sin(\alpha+\beta)}{\sin\alpha\cos\beta - \cos\alpha\sin\beta \cos\theta}\,(\cos\alpha \cos\theta, \cos\alpha\sin\theta,\sin\alpha)$$
A tangent vector $P'$ is proportional to the derivative of $P$ with respect to $\theta$; in particular, we can take
$$P' = (
\sin\alpha\cos\beta\sin\theta,
\cos\alpha\sin\beta-\sin\alpha\cos\beta\cos\theta,
\sin\alpha \sin\beta \sin\theta)$$
Then, the angle $\psi$ between the generator $OP$ and tangent vector $P'$ satisfies
$$\cos\psi = \frac{P\cdot P'}{|P|\,|P'|}$$
We have
$$\begin{align}
P\phantom{^\prime}\cdot P' &= \frac{v \sin\beta \sin(\alpha+\beta) \sin\theta}{
\sin\alpha\cos\beta - \cos\alpha\sin\beta\cos\theta} \\[4pt]
|P\phantom{^\prime}|^2 = P\phantom{^\prime}\cdot P\phantom{^\prime} &= \frac{v^2 \sin^2(\alpha+\beta)}{(\sin\alpha\cos\beta
- \cos\alpha\sin\beta\cos\theta)^2} \\[4pt]
|P'|^2 = P'\cdot P' &= (\sin\alpha\cos\beta - \cos\alpha\sin\beta\cos\theta)^2
+ \sin^2\beta \sin^2\theta
\end{align}$$
Thus,
$$\begin{align} \cos\psi &= \phantom{\pm}\frac{\sin\beta \sin\theta}{\sqrt{(\sin\alpha\cos\beta - \cos\alpha\sin\beta\cos\theta)^2 + \sin^2\beta \sin^2\theta}} \\[4pt] \sin\psi &= \pm\frac{\sin\alpha\cos\beta - \cos\alpha\sin\beta\cos\theta}{\sqrt{(\sin\alpha\cos\beta - \cos\alpha\sin\beta\cos\theta)^2 + \sin^2\beta \sin^2\theta}} \\[4pt] \cot\psi &= \pm\frac{\sin\beta\sin\theta}{\sin\alpha\cos\beta - \cos\alpha\sin\beta\cos\theta} = \pm\frac{\sin\theta}{\sin\alpha\cot\beta - \cos\alpha\cos\theta} \\[4pt] \end{align} $$
for appropriate choices of sign. $\square$
(I was hoping for something a little more elegant.)
Incidentally, one can show that, in general, the eccentricity of the conic is given by $e = \sin\beta/\sin\alpha$, but this observation doesn't seem to make the formulas appreciably better.
For the "standard" cone with $\alpha=\pi/4$, we have $$\cot\psi = \pm\frac{\sqrt{2}\,\sin\theta}{\cot\beta-\cos\theta}$$
For any cone cut by a horizontal plane, $\beta=0$ (giving a circle), we have $$\cot\psi = 0$$ so that $\psi$ is constantly $\pi/2$, which is geometrically obvious (as noted by OP).
For the standard cone cut by a vertical plane, $\beta=\pi/2$ (giving a rectangular hyperbola), we have $$\cot\psi = \pm \sqrt{2}\tan\theta \quad\to\quad \cot\theta\cot\psi = \pm\sqrt{2}$$ The reader can verify that this agrees with @Aretino's formula, with the condition $a=b$.
For the standard cone cut by a comparably-inclined plane, $\beta=\alpha=\pi/4$ (giving a parabola), we have $$\cot\psi = \pm\frac{\sqrt{2}\,\sin\theta}{1-\cos\theta} = \pm\sqrt{2}\,\cot\frac12\theta$$
It's true (except possibly for edge cases).
It is fairly obvious if you draw a diagram with two unit circles - one at the origin with vectors $\mathbf{a}$ and $\mathbf{b}$ in it, and the other unit circle centred at $\mathbf{a}+\mathbf{b}$ and which contains the vector $-\mathbf{c}$. Vectors $\mathbf{a}$ and $\mathbf{b}$ will then point to the intersections of the unit circles.
For $|\mathbf{a} + \mathbf{b} - \mathbf{c}| < 1$ to hold, the head of $-\mathbf{c}$ must lie inside the first unit circle, and this happens exactly when it lies on the circular arc between the two intersection points. Therefore by symmetry of the paralellogram the same is true for $\mathbf{c}$ when placed at the origin. Those intersection points are pointed at by $\mathbf{a}$ and $\mathbf{b}$, so $|\mathbf{a} + \mathbf{b} - \mathbf{c}| < 1$ iff $\mathbf{c}$ lies strictly between $\mathbf{a}$ and $\mathbf{b}$.
Note that you used a $\le$ sign, but you'll have to think about whether that is what you want or if strict inequality is better. Does $\mathbf{a}$ itself lie between $\mathbf{a}$ and $\mathbf{b}$? And in the case of $\mathbf{a}=-\mathbf{b}$, does every unit vector lie between them or none at all?
Best Answer
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.