Subject:
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Re: Yet another math problem
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Newsgroups:
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lugnet.off-topic.geek
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Date:
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Mon, 11 Nov 2002 19:59:40 GMT
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Viewed:
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381 times
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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)
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Message has 2 Replies:
 | | Re: Yet another math problem
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| (...) 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)
| | | Re: Yet another math problem
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| (...) Hey, that's super! And I finally found an online reference at the U of Georgia site, so I can probably handle these in the future. Thanks for the clear explanation and the solution. Dave! (22 years ago, 11-Nov-02, to lugnet.off-topic.geek)
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