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Topic:
Inverse of a conic with regard to the centre is an oval of Cassini
Replies:
6
Last Post:
Jul 28, 2011 7:33 AM




Re: Inverse of a conic with regard to the centre is an oval of Cassini
Posted:
Jul 27, 2011 8:08 PM


In article <4e5b37473f844764b0c33f48e796b52f@e8g2000vbw.googlegroups.com>, Vivian DeRoth <vivianderoth@gmail.com> wrote: >I came across the following link: >http://www.archive.org/stream/differentialcalc00edwauoft/differentialcalc00edwauoft_djvu.txt > which included the following few lines: > >48. Show that the inverse of a conic with regard to the focus >is a Lima^on (Equation r = a + b cos 0), which becomes a cardi >oide if the conic be a parabola. > >49. Show that the inverse of a conic with regard to the >centre is an oval of Cassini (Equation r 2 = + 6cos2#), which >becomes a Lemniscate of Bernoulli if the conic be a rectangular >hyperbola. > > I thought i read somewhere that the inverse of a Cassinian was a >Cassinian  i think the word for this property is anagammatic or >something like that. > >Is that true what is said in 49. ie that the 'inverse of a conic with >regard to the >centre is an oval of Cassini'. If so can anyone point me towards >sources that would show how this is.
I did the inversion for an ellipse, and I get the curve below. It is hard to read the equation you have above (I assume that some characters did not copy and paste well), but it looks similar. However, the curve does not seem to be an Oval of Cassini as described on <http://en.wikipedia.org/wiki/Cassini_oval>.
Let us start with the ellipse
x^2 y^2  +  = 1 a^2 b^2
If we invert the points (x > x/r^2, y > y/r^2), we get the curve
x^2 y^2  +  = r^4 a^2 b^2
Divide both sides by r^2 to get
cos^2(?) sin^2(?)  +  = r^2 a^2 b^2
1+cos(2?) 1cos(2?)  +  = r^2 2 a^2 2 b^2
1 1 1 1 1 1  (  +  ) +  (    ) cos(2?) = r^2 2 a^2 b^2 2 a^2 b^2
Rob Johnson <rob@trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font



