Conic Sections

\[ Ax^2+Bxy+Cy^2+Dx+Ey+F=0. \]

\[ \mathbf{x}^{\mathrm{T}}\mathbf{A}\mathbf{x}=0 \] where \[ \mathbf{x}=\begin{bmatrix}x\\ y \\1\end{bmatrix}\quad \mathbf{A}=\begin{bmatrix}A&B/2&D/2\\B/2&C&E/2\\D/2&E/2&F\end{bmatrix} \]

• Extraordinary Conics: The Most Difficult Math Problem I Ever Solved - YouTube
  • A set with size five of points and lines can define a conic section.
• \[ \mathbf{x}=\mathbf{a}\sin(\theta)+\mathbf{b}\cos(\theta)+\mathbf{c} \] or \[ \mathbf{x}=\mathbf{a}\sinh(\theta)\pm\mathbf{b}\cosh(\theta)+\mathbf{c} \] for any non-colinear vectors \(\mathbf{a}, \mathbf{b}\).
• Area \[ A=\pi|\mathbf{a}\times\mathbf{b}| \]
• \(c^2\) Invariant \[ c^2=\mathbf{a}^2+\mathbf{b}^2 \]

• Interiority Test

  \begin{equation*} |(\mathbf{p}-\mathbf{c})\times \mathbf{a}|^2+|(\mathbf{p}-\mathbf{c})\times\mathbf{b}|^2< |\mathbf{a}\times\mathbf{b}|^2 \end{equation*}
• Tangency Test

\begin{equation*} |\mathbf{r}\times\mathbf{a}|^2+|\mathbf{r}\times\mathbf{b}|^2=|\mathbf{r}\times(\mathbf{p}-\mathbf{c})|^2 \end{equation*}

where \(\mathbf{p}\) is the point of tangency and \(\mathbf{r}\) is the direction in which the tangent line is.

\begin{equation*} |v_1, v_5, v_4||v_3, v_4, v||v_2, v, v_5||v_1, v_2, v_3| - |v_1, v_5, v_2||v_3, v_4, v_1||v_2, v, v_3||v_4, v_5, v| = 0. \end{equation*}

• Create any conic section with five points from an equation hiding in Pappus' …

Given 9 points(18 degrees of freedom) within the affine space \(v_1, v_2, \dots, v_9\) with 9 collinearity conditions(removes 8 degrees of freedom since the last condition is dependent to the previous), by the Pappus's hexagon theorem:

\begin{equation*} |v_7, v_8, v_9| = \frac{|v_1, v_5, v_4||v_3, v_4, v_6||v_2, v_6, v_5||v_1, v_2, v_3| - |v_1, v_5, v_2||v_3, v_4, v_1||v_2, v_6, v_3||v_4, v_5, v_6|}{|v_1, v_5, v_2-v_4||v_3, v_4, v_1-v_6||v_2, v_6, v_3-v_5|} = 0. \end{equation*}

Let any of the 6 points, say \(v_6\) be the variable:

\begin{equation*} v = \begin{bmatrix} x \\ y \\ 1 \end{bmatrix} \end{equation*}

Yielding the equation of a conic section that contains h \(v_1, v_2, \dots, v_5\):

\begin{equation*} |v_1, v_5, v_4||v_3, v_4, v||v_2, v, v_5||v_1, v_2, v_3| - |v_1, v_5, v_2||v_3, v_4, v_1||v_2, v, v_3||v_4, v_5, v| = 0. \end{equation*}

• This holds due to the Pascal's theorem.
• Braikenridge-Maclaurin theorem guerantees the existence of the conic containing five points.

\begin{equation*} \begin{vmatrix} x^2 & xy & y^2 & x & y & 1 \\ x_1^2 &x_1y_1 & y_1^2 & x_1 & y_1 & 1 \\ x_2^2 &x_2y_2 & y_2^2 & x_2 & y_2 & 1 \\ x_3^2 &x_3y_3 & y_3^2 & x_3 & y_3 & 1 \\ x_4^2 &x_4y_4 & y_4^2 & x_4 & y_4 & 1 \\ x_5^2 &x_5y_5 & y_5^2 & x_5 & y_5 & 1 \\ \end{vmatrix} = 0. \end{equation*}

For an ellipsoid centered at the origin, the ellipsoid corresponds to a quadratic form

\begin{equation*} \mathbf{x}^\top \mathbf{A} \mathbf{x} = 1 \end{equation*}

where \( \mathbf{x} \) is the coordinate vector.

The vectors that represent conjugate axes \( \{ v_i \} \) are the set of vectors that satisfy:

\begin{equation*} \mathbf{v}_i^{\top} \mathbf{A}\mathbf{v}_j = \delta_{ij}. \end{equation*}

• \[ Ax^2+2Bxy+Cy^2+2Dxz+2Eyz+Fz^2=0 \]

\begin{align*} &[((x-x_{0}) \cos(\phi)+(y-y_{0})\sin(\phi))\cos(\theta)-(z-z_{0}) \sin(\theta)]^2\\ +&[-(x-x_{0}) \sin(\phi)+(y-y_{0}) \cos(\phi)]^2\\ -&[((x-x_{0}) \cos(\phi)+(y-y_{0}) \sin(\phi)) \sin(\theta)+(z-z_{0}) \cos(\theta)]^2=0 \end{align*}

• Standard form
  • \( (x-x_0)^2+(y-y_0)^2=r^2 \)
• General form
  • \( ax^2+by^2+cx+dy+e=0 \) where \(a=b\).
• Circle with diameter \(\rm\overline{AB}\)
  • \( (x-a_x)(x-b_x)+(y-a_y)(y-b_y)=0 \)
  • In vector notation, \( (\mathbf{x}-\mathbf{a})\cdot(\mathbf{x}-\mathbf{b})=0 \) which describes all the points such that vector from \( \rm A \)and vector from \( \rm B \)are perpendicular.
    • By Thales's theorem
• Vector form
  • \( (\mathbf{x}-\mathbf{x}_0)\cdot(\mathbf{x}-\mathbf{x}_0)=r^2 \)

• This task is usually a chore – not anymore! - YouTube
• A circle is completely defined by any three non-colinear points \(\rm A,B, C\).
• From the equation of circle with diameter \(\rm\overline{AB}\), \(f(x,y)=0\) and the equation of line that goes through \(\rm A,B\), \(g(x,y)=0\), one can find a real number \(\alpha\) such that \(f(x,y)=\alpha g(x,y)\) contains all three points.

• Points with a fixed ratio of distances from two points form a circle.

• For an ellipse parameterized by \((\cos\theta, a\sin\theta)\), at every point:
  • \[ r^2 = 1+e^2y^2. \]
• Apsis
  • \[ a = \frac{r_{\rm ap} + r_{\rm per}}{2} \]
    • \[ c = \frac{r_{\rm ap} - r_{\rm per}}{2} \]
  • \[ b = \sqrt{ r_{\rm ap}r_{\rm per} } \]
  • \[ \ell = \left( \frac{r_{\rm ap}^{-1} + r_{\rm per}^{-1}}{2}\right)^{-1} \]
  • \[ e = \frac{r_{\rm ap} - r_{\rm per}}{r_{\rm ap} + r_{\rm per}} \]
• Tangential Angle
  • \[ \sin\phi = \frac{b}{\sqrt{r(2a-r)}} \]
    • where \(\phi\) is the angle of the tangent line with respect to one of the segments from the foci, and \(r\) is the length of one of the segments.
  • This can be obtained from the :
    • \[ \begin{align*} (2c)^2 &= r^2 + s^2 - 2rs\cos(\pi - 2\phi)\\ \implies (2c)^2 &= (r+s)^2 - 4rs\sin^{\cdot 2}\phi \\ \implies \sin\phi &= \sqrt{\frac{(2a)^2 - (2c)^2}{4rs}} \end{align*} \]
  • The same form can also be derived independently from the angular momentum conservation and energy conservation.
  • Proof of Kepler's Elliptical Orbit Law

• \[ x^2 - y^2 = 1 \]

• \[ y = \frac{1}{x} \iff 1 = \frac{1}{xy} \iff xy = 1 \]

• Scaling in one of the axes: \[ y = \frac{1}{(kx)} \iff ky = \frac{1}{x} \] is equivalent to scaling in both axes by the square root of the factor: \[ \sqrt{k}y = \frac{1}{\sqrt{k} x}. \]

• It can be thought of squishing in the direction of one of the asymptote, and stretching it the direction of the other asymptote by the same factor, .
• When The wedges under the hyperbola gets hyperbolically rotated, the total area is preserved, which is similar to the rotation of a circle.
• It can be represented by a symmetric matrix with determinant of one.

See hyperbolic functions

• excalidraw:parabola.excalidraw
• The points \( (a, a^2), (b, b^2), (0,ab) \) are colinear.
• What's Behind the Parabola? (#SoME3) - YouTube

• The affine # Algebraic Variety in the form of \[ \mathbf{x}^{\rm T}\mathbf{Q}\mathbf{x} + \mathbf{P}\mathbf{x} + R = 0. \]