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January 11th, 2013, 07:22 PM   #1
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Radical And trigonometry simplification

The following expression pertains to the fact that the sum of the distances from the foci to any point on an ellipse is 2a, where a is the radius along the major axis.



I have verified this equation for various values of a, b, and theta, so I know that it is correct. I don't know how to verify it algebraically, though. Can anyone show me how to do it? I've tried squaring it, but it just got really nasty.
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January 11th, 2013, 07:28 PM   #2
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Re: Radical And trigonometry simplification

Did you try arranging the equation with a radical on either side, then squaring twice? I believe that's what's usually done with ellipses and hyperbolas.
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January 11th, 2013, 07:42 PM   #3
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Re: Radical And trigonometry simplification

Sounds like you're suggesting something like



I expanded both the left and right sides in maple, and it seems to just generate more complex terms that are difficult to cancel :\ The issues come from the interior terms of the expansion that do not eliminate the square roots.
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January 11th, 2013, 07:50 PM   #4
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Re: Radical And trigonometry simplification

Nevermind haha, I think I get what to do now. This reference seems to show exactly what I'm trying to do

http://en.wikipedia.org/wiki/Proofs_inv ... he_ellipse

Thanks for the help, and you're right about the double squaring!
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January 11th, 2013, 07:53 PM   #5
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Re: Radical And trigonometry simplification

Yes, you've got radicals under radicals...perhaps if you rewrite the expressions a different way. For convenience, let F be half the interfocal distance, where:

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