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Tangent for Parabolas
Posted:
Nov 11, 1996 11:03 PM


I'm wondering about the following:
Suppose i have two parabolas, with the standard equations
(x  h)^2 = 4p(y  k) > (1) (y  k)^2 = 4p(x  h) > (2)
And am asked to find the slope of the tangent at the point (x0, y0).
I then tried to differentiated both, but only found success for (1).
I found that the derivative for (1) to be
m = (x0  h) / 2p.
(i used m because it stands for slope) and found it consistent with a graphing program.
For (2), i could not find an answer, but i came out with this
m = sqr(p) / x  h
However, i found it inconsistent with the same graphing program. Perhaps the graphing program's wrong, maybe not.
++  Is there a derivative for (2) ?  ++
As a final question, providing that my (1) derivative is correct, how come it doesn't involve the k, or y0? My guess is that somehow the x0 determines the y0, so the y0 is not need. I don't know about the k, though.
Thanks, Scott



