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Polar equation to cartesian

  • #1
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i have this equation: [tex]r=\sqrt{1+sin2\theta}[/tex]
and am to convert to cartesian equation and from the equation see that it consists of two circles and directly note the radii of the cirlcles from the equation.

so far i have manipulated it and gotten:

[tex]x^6 + 3x^4 y^2 + 3 x^2 y^4 + y^6 - x^4 -6x^2 y^2 - 4x^3 y -y^4 - 4y^3 x=0[/tex]

any suggestions on how to get it into the required form?
thanks
 

Answers and Replies

  • #2
Tide
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HINT: [tex]\sin \alpha = \cos \left (\frac {\pi}{2} - \alpha\right)[/tex]
 
  • #3
benorin
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It should be a fourth order equation in x and y. I get:

[tex]\left[ \left(x+\frac{1}{2}\right) ^2 + \left(y+\frac{1}{2}\right) ^2 -\frac{1}{2}\right] \left[ \left(x-\frac{1}{2}\right) ^2 + \left(y-\frac{1}{2}\right) ^2 -\frac{1}{2}\right] =0[/tex]
 
  • #4
dextercioby
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Well, i get something like

[tex] x^2 +y^2 =1+\frac{2xy}{x^2 +y^2} [/tex],

from which the conclusion follows easily.

Daniel.
 
  • #5
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dextercioby said:
Well, i get something like
[tex] x^2 +y^2 =1+\frac{2xy}{x^2 +y^2} [/tex],
from which the conclusion follows easily.
Daniel.
yes, i too found this but can you explain how the conclusion follows easily from this?
 
  • #6
arildno
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For starters, rewrite it in the form:
[tex](x^{2}+y^{2})^{2}=(x+y)^{2}[/tex]
 
  • #7
arildno
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benorin said:
then factor it using difference of squares
That was to be my next hint..:frown:

Then, the grand finale would have been to refer to your earlier solution to this problem. I won't do that now.
 
  • #8
benorin
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There :rolleyes: , or rather not there: all better?
 

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