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Subject: 
Re: Yet another math problem
Newsgroups: 
lugnet.off-topic.geek
Date: 
Mon, 11 Nov 2002 20:10:01 GMT
Viewed: 
407 times
  
In lugnet.off-topic.geek, Adrian Drake writes:
In lugnet.off-topic.geek, Jeff Jardine writes:
In lugnet.off-topic.geek, Dave Schuler writes:

I have five points and am trying to define the parabola that contains them
(if such exists).  The points are:

(0,0) (which is also the vertex)
(14,100)
(-14,100)
(30,180)
(-30,180)


Hmm.  Well, now that I think of it, the vertex could be x=0 with y as an
unknown.  Does that change anything?


Yes, it does.  If you just have the last four points to work with, then you
definitely have an upward-turned parabola, symmetrical about the y-axis.

Here's what you have to do to find the formula for a parabola, given three
points:

y = ax^2 + bx + c

plug in your three points:
180 = a(30)^2 + b(30) + c
180 = a(-30)^2 + b(-30) + c
100 = a(14)^2 + b(14) + c

You have three equations with three unknowns.  From the first two, you can
see that b has to be 0, so we're left with:

180 = 900a + c
100 = 196a + c

so, a = 5/44, and c = 855/11, so the vertex intercepts the y-axis at 855/11
(about 77.7)

Your solution is y = (5/44)x^2 + 855/11

Jeff J
(this time I double checked everything before clicking 'post' like a good
geek should)

Sorry, but this doesn't go through 0,0.  Solving x=0, y=0 gives 0=855/11,
which is clearly not true, so this isn't the equation he's looking for either.


Oops.  Hang on there Adrian, read for content.

The five points as given in the initial problem (including 0,0) will not
solve to a simple parabolic equation.  Eliminating 0,0 does give us the
solution specified above.  Bleh.

Adrian



Message is in Reply To:
  Re: Yet another math problem
 
(...) Sorry, but this doesn't go through 0,0. Solving x=0, y=0 gives 0=855/11, which is clearly not true, so this isn't the equation he's looking for either. Adrian -- www.brickfrenzy.com (22 years ago, 11-Nov-02, to lugnet.off-topic.geek)

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