Derivation of the Equation of an Ellipse
Date: 06/23/99 at 19:04:52 From: Rob Walsman Subject: Derivation of equation of an ellipse I have been trying to derive the equation of an ellipse from its definition (the locus of points having a fixed sum of distances to the two foci - sorry for the loose definition). My algebra gets very complex trying to eliminate radicals. Could you point me to a source where I can see a good derivation? Thanks, Rob Walsman
Date: 06/24/99 at 15:43:52 From: Doctor Peterson Subject: Re: Derivation of equation of an ellipse Hi, Rob. The derivation really isn't too hard, but you have to know how to handle an equation containing the sum of two radicals. The trick is to move one radical to the other side and square; this leaves you with only one radical. Then you can rearrange the equation so this radical is alone on one side, then square again to eliminate the radical entirely. Here's how it goes: (x,y)********+******** ****+*___ | \_ ****** *** / \____| \__a *** ** / \b___ \_ ** * / | \____\_ * *(-f,0)/ f | f \_\_(f,0) * *-----+--------------+-------------+------* * a | a * * | * ** |b ** *** | *** ****** | ****** ***************** We want the sum of the distances from (-f,0) to (x,y) and from (x,y) to (f,0) to be 2a, where a is the horizontal semiaxis; the vertical semiaxis will be b, where b^2 + f^2 = a^2. The equation is sqrt[(x+f)^2 + y^2] + sqrt[(x-f)^2 + y^2] = 2a Move the second root to the right and square: sqrt[(x+f)^2 + y^2] = 2a - sqrt[(x-f)^2 + y^2] (x+f)^2 + y^2 = 4a^2 - 4a sqrt[(x-f)^2 + y^2] + (x-f)^2 + y^2 x^2 + 2fx + f^2 + y^2 = 4a^2 + x^2 - 2fx + f^2 + y^2 - 4a sqrt[(x-f)^2 + y^2] Move the root to the left side by itself, and divide by 4: 4a sqrt[(x-f)^2 + y^2] = 4a^2 - 4fx a sqrt[(x-f)^2 + y^2] = a^2 - fx Square both sides again: a^2 (x^2 - 2fx + f^2 + y^2) = a^4 - 2a^2 fx + f^2 x^2 a^2 x^2 - 2a^2 fx + a^2 f^2 + a^2 y^2 = a^4 - 2a^2 fx + f^2 x^2 (a^2 - f^2)x^2 + a^2 y^2 = a^4 - a^2 f^2 Divide by a^2(a^2 - f^2): x^2 y^2 --- + --------- = 1 a^2 a^2 - f^2 Write b^2 = a^2 - f^2, and we're done: x^2 y^2 --- + --- = 1 a^2 b^2 It's always fun to see how much cancels out when you're on the right track! - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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