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November 25th, 2009, 10:07 PM   #1
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Find equation of the common tangent to the parabolas

Find equation of the common tangent to the parabolas y^2 = 4ax and x^2 = 4by
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November 26th, 2009, 01:22 PM   #2
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The tangent to the parabola y = 4ax at the point (at, 2at) has equation ty = x + at.
The tangent to the parabola x = 4by at the point (2bu, bu) has equation ux = y + bu.
If the above tangents have the same equation, tu = 1 and bu = -at, so u = -a/b.
Hence u = ?(-a/b) and t = ?(-b/a), where "?" specifies the real cube root.
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December 1st, 2009, 09:19 PM   #3
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Re: I couldnt understand the last 2 lines

Quote:
Originally Posted by skipjack
If the above tangents have the same equation, tu = 1 and bu = -at, so u = -a/b.
Hence u = ?(-a/b) and t = ?(-b/a), where "?" specifies the real cube root.

Please could u explain me how did u get tu = 1 , bu^2 = -at from the eg. of tangents....
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December 1st, 2009, 11:13 PM   #4
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I multiplied the second equation by t, then compared it to the first equation. One term is already the same, and the equations I gave follow if the other two terms are the same.
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