2

For example,

  1. Circle ($\frac{x^2}{a^2}+\frac{y^2}{a^2}=1$) : $$ \left\{ \begin{array}{c} x\to a\cos \theta \\ y\to a \sin \theta \\ \end{array} \right. $$
  2. Oval ($\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$) : $$ \left\{ \begin{array}{c} x\to a\cos \theta \\ y\to b \sin \theta \\ \end{array} \right.$$

Any other equations containing $x$,$y$,$x^2$, and $y^2$ can be substituted to trigonometric equations like this, with some translations.

However, I can't find out any way to do this on equations containing $xy$ terms.

For example, $$Ax^2 + By^2 + Cxy + Dx + Ey + F = 0$$, where $C \ne 0.$ How can I do that? Thanks!

KYHSGeekCode
  • 777
  • Are you assuming that the conic section is an "oval" (usually called ellipse), that is $C^2-4AB<0$? A strategy could be: 1) Use a rotation of conic sections to eliminate $C$ and express the conic section in standard form in $x'y'$-coordinates. 2) Use your formula for ovals to express $x'$ and $y'$ using an angle. 3) Use the relation between $(x,y)$ and $(x',y')$ to find a parametrisation of $x$ and $y$ using an angle. This is straightforward in theory, more difficult in practice (because of the 6 parameters $A,B,\dots,F$) – Taladris Oct 08 '18 at 04:20
  • Note: when writing $x=a\cos(\theta)$ and $y=b\sin(\theta)$ with $a\neq b$, the parameter $\theta$ is not the angle that appears in the polar coordinates of $(x,y)$. – Taladris Oct 08 '18 at 04:21
  • @Taladris Yes, $C^2-4AB<0$. Thank you for your suggestion. Do you think there will be no other ways to do this substitution than the method you mentioned? Also thank you for your pointing out that $\theta$ is not an angle when $a\ne b$. – KYHSGeekCode Oct 08 '18 at 04:31
  • Another idea: find the center $\Omega(a,b)$ of the ellipse, and consider the ray emanating from it and making an angle $\theta$ with the horizontal axis. This ray intersects the ellipse in exactly one point $P$. If $r$ is the distance between $P$ and $\Omega$, then $P(x,y)$ can be described as $x=a+r\cos(\theta)$, $y=b+r\sin(\theta)$. Note that $r$ is a function of $\theta$. – Taladris Oct 08 '18 at 04:41
  • Please refer to another answer here. – Ng Chung Tak Oct 09 '18 at 14:22
  • @Taladris Thanks for your various suggestions. I may try that second method in near future. – KYHSGeekCode Oct 09 '18 at 15:41
  • @NgChungTak Thanks, I will deeply read that when I have time to. – KYHSGeekCode Oct 09 '18 at 15:45

1 Answers1

2

I will use the slightly more standard $ax^2+2bxy+cy^2+2dx+2ey+f=0,$ assumed to be an ellipse. Its center is $(\frac{cd-eb}{b^2-ac},\frac{ae-bd}{b^2-ac})$ and translating this to the origin we get

$$ax'^2+2\,bx'y'+cy'^2={\frac {-cd^2+acf-ae^2-fb^2+2deb}{{b}^{2}-ac}}$$ Rotating through the angle $\theta'=\arctan{(\frac{-c+a+\sqrt{c^2-2ac+a^2+4b^2}}{2b})}$ we get $${\frac {-4\,c{b}^{2}-2\,{b}^{2}\sqrt {{c}^{ 2}-2\,ac+{a}^{2}+4\,{b}^{2}}-{c}^{3}+2\,a{c}^{2}-{c}^{2}\sqrt {{c}^{2} -2\,ac+{a}^{2}+4\,{b}^{2}}-c{a}^{2}+ac\sqrt {{c}^{2}-2\,ac+{a}^{2}+4\, {b}^{2}}}{-4\,{b}^{2}-{c}^{2}+2\,ac-c\sqrt {{c}^{2}-2\,ac+{a}^{2}+4\,{ b}^{2}}-{a}^{2}+a\sqrt {{c}^{2}-2\,ac+{a}^{2}+4\,{b}^{2}}}}x''^2+{\frac {4\,c{b}^{2}-2\,{b}^{2}\sqrt {{c}^{2}-2 \,ac+{a}^{2}+4\,{b}^{2}}+{c}^{3}-2\,a{c}^{2}-{c}^{2}\sqrt {{c}^{2}-2\, ac+{a}^{2}+4\,{b}^{2}}+c{a}^{2}+ac\sqrt {{c}^{2}-2\,ac+{a}^{2}+4\,{b}^ {2}}}{4\,{b}^{2}+{c}^{2}-2\,ac-c\sqrt {{c}^{2}-2\,ac+{a}^{2}+4\,{b}^{2 }}+{a}^{2}+a\sqrt {{c}^{2}-2\,ac+{a}^{2}+4\,{b}^{2}}}}y''^2 -{\frac {-cd^2+acf-ae^2-fb^2+2deb}{{b}^{2}-ac}}=0$$ This can now be put into the form $\frac{x''^2}{a''^2}+\frac{y''^2}{b''^2}=1.$

In an example: $-5x^2+4xy+7x-4y^2-4y+24=0$ is an ellipse with center $(\frac58,-\frac3{16}).$ The method yields $-{\frac {16}{425}}\,{\frac {{x''}^{2} \left( 68-4\,\sqrt {17} \right) }{ -17-\sqrt {17}}}-{\frac {16}{425}}\,{\frac {{y''}^{2} \left( -68-4\, \sqrt {17} \right) }{17-\sqrt {17}}}-1$ or $a''={\frac {85}{32}}+{\frac {5}{32}}\,\sqrt {17}$ $b''={\frac {85}{32}}-{\frac {5}{32}}\,\sqrt {17}$.

Going back through the substitutions, $x''=\cos{\theta'}x'+\sin{\theta'}y',y''=-\sin{\theta'}x'+\cos{\theta'}y'$ and $x'=x-\frac58$,$y'=y+\frac3{16}$ we get $$\begin{array}{cc} x=&\frac58-{\frac {5}{16}}\,\sqrt {2}\sin \left( t \right) \sqrt {17-\sqrt { 17}}+{\frac {5}{16}}\,\cos \left( t \right) \sqrt {34+2\,\sqrt {17}} \\y=& -\frac3{16}+{\frac {5}{1088}}\,\sqrt {17} \left( -\sqrt {2}\sqrt {17}\sin \left( t \right) \sqrt {17-\sqrt {17}}+17\,\sqrt {2}\sin \left( t \right) \sqrt {17-\sqrt {17}}+\cos \left( t \right) \sqrt {34+2\, \sqrt {17}}\sqrt {17}+17\,\cos \left( t \right) \sqrt {34+2\,\sqrt {17 }} \right). \end{array}$$

Jan-Magnus Økland
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