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s1119 - x 2 y 2 1-e 2 = p 2 1-e 2 ± p e 1-e 2 ² 2 = p...

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Solution for assignment 19 set 6. Rewritten and using cos θ = x/r , the initial equation becomes p = r (1 + e cos θ ) = r (1 + e x/r ) = r + e x or r = p - e x . Squaring both sides x 2 + y 2 = p 2 - 2 p e x + e 2 x 2 . Bringing all terms with x or y to one side, x 2 (1 - e 2 ) + 2 p e x + y 2 = p 2 , x 2 + 2 p e 1 - e 2 x + y 2 1 - e 2 = p 2 1 - e 2 , According to the recipe for completion of the square we substitute x 0 = x +
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Unformatted text preview: x 2 + y 2 1-e 2 = p 2 1-e 2 + ± p e 1-e 2 ² 2 = p 2 (1-e 2 ) + p 2 e 2 (1-e 2 ) 2 = p 2 (1-e 2 ) 2 , x 2 (1-e 2 ) 2 p 2 + y 2 1-e 2 p 2 = 1 . With the deFnitions a = p 1-e 2 and b = p √ 1-e 2 this reads x a ! 2 + ± y b ² 2 = 1 ....
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