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 Pre-Calculus and Calculus
 derivation of equation for ellipse
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newton
Junior Member

USA
4 Posts

Posted - 10/16/2007 :  11:26:41  Show Profile
I am trying to simplify the following equation:

((x+c)+y) + ((x-c)+y) = 2a

to get: x/a + y/(a - c) = 1.
I brought the second radical to the right side of the equation, squared and start simplifying:

(x+c)+y = 4a-4a ((x-c)+y) + (x-c)+y
(x+c)+y) - (x-c)-y = 4a - 4a((x-c)+y)
(x+c) - (x-c) = 4a-4a((x-c)+y)
4cx = 4a-4a((x-c)+y)
cx = a-a((x-c)+y)
cx - a = -a((x-c)+y)

How do I proceed from here, to get what I want above?

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peppers18
Advanced Member

Canada
124 Posts

Posted - 10/16/2007 :  12:01:58  Show Profile
try this website http://en.wikipedia.org/wiki/Ellipse/Proofs
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Subhotosh Khan
Advanced Member

USA
9117 Posts

Posted - 10/16/2007 :  12:19:55  Show Profile
quote:
Originally posted by newton

I am trying to simplify the following equation:

((x+c)+y) + ((x-c)+y) = 2a

to get: x/a + y/(a - c) = 1.
I brought the second radical to the right side of the equation, squared and start simplifying:

(x+c)+y = 4a-4a ((x-c)+y) + (x-c)+y
(x+c)+y) - (x-c)-y = 4a - 4a((x-c)+y)
(x+c) - (x-c) = 4a-4a((x-c)+y)
4cx = 4a-4a((x-c)+y)
cx = a-a((x-c)+y)
cx - a = -a((x-c)+y)<---Square both sides again and simplify

However I get equation of hyperbola from there - check your signs

How do I proceed from here, to get what I want above?




Edited by - Subhotosh Khan on 10/16/2007 12:32:53
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newton
Junior Member

USA
4 Posts

Posted - 10/16/2007 :  15:31:49  Show Profile
To peppers18.
I checked the website.
That is exactly what I was looking for.
Thank you.
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