The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » Math Topics » alt.math.undergrad

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Tangent for Parabolas
Replies: 2   Last Post: Nov 13, 1996 10:26 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Scott Phung

Posts: 32
Registered: 12/6/04
Tangent for Parabolas
Posted: Nov 11, 1996 11:03 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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.


Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.