Loci Conics Parabola eqn in standard form


by aisha
Tags: conics, form, loci, parabola, standard
aisha
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#1
Aug24-05, 08:21 PM
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I need to find the equation of a parabola in standard form if the domain of an arch under a bridge is {-50 <= x <= 50} and the range is {0<=y<=20}

The span of the arch is 100 metres and the height is 20 metres.

standard form of a parabola is

[tex] (x-h)^2=4p(y-k) [/tex]

i know the vertex is (0,20) so the equation should look like [tex] x^2=4p (y-20) [/tex] so far but im not sure what the focus is or how to determine it plz help me out
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Hurkyl
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#2
Aug24-05, 08:24 PM
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Do you know of any other points on your parabola?
HallsofIvy
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#3
Aug25-05, 09:18 AM
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Quote Quote by aisha
I need to find the equation of a parabola in standard form if the domain of an arch under a bridge is {-50 <= x <= 50} and the range is {0<=y<=20}

The span of the arch is 100 metres and the height is 20 metres.

standard form of a parabola is

[tex] (x-h)^2=4p(y-k) [/tex]

i know the vertex is (0,20) so the equation should look like [tex] x^2=4p (y-20) [/tex] so far but im not sure what the focus is or how to determine it plz help me out
What are the coordinates of the ends of the bridge?

aisha
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#4
Aug25-05, 05:01 PM
P: 587

Loci Conics Parabola eqn in standard form


Well yes I think I do know two more points

(0,-50) and (0,50) what do I do to complete my standard form equation?

Do I sub in one of the points into [tex] x^2=4p (y-20) [/tex] and then solve for p?

I dont think i'm doing this right
[tex] x^2=4p (y-20) [/tex]
[tex] 0^2=4p(50-20) [/tex]
[tex] 0=4p(30) [/tex]
[tex] 0=120p [/tex]

now what?? I dont know what to do please help me!
HallsofIvy
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#5
Aug25-05, 07:07 PM
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Okay, if (0,50) and (0,-50) have to satisfy the equation, how about you put x= 0, y= 50 into the equation and see what happens?
aisha
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#6
Aug25-05, 08:32 PM
P: 587
that's what I have done in the last post....I'm still not sure what to do? plzzzz help!!
EnumaElish
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#7
Aug25-05, 11:20 PM
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Quote Quote by aisha
but im not sure what the focus is
Did you mistype "locus" as "focus"?

Let's see: (0, 20) is the vertex. (-50, 0) and (50, 0) are the endpoints. If I understand what the span is, then the endpoints should have y = 0 instead of x = 0. (That is, I believe that you interchanged the x value and the y value of the two endpoints by mistake.)

You have [tex](x-h)^2=4p(y-k)[/tex] with 3 parameters to calculate: h, k and p.

You have the 3 points and should be able to solve the following 3 equations for h, k and p:

[tex] (0-h)^2=4p(20-k) \ ... \text{ Eq. (1)}[/tex]
[tex] (-50-h)^2=4p(0-k) \ ... \text{ Eq. (2)}[/tex]
[tex] (50-h)^2=4p(0-k) \ ... \text{ Eq. (3)}[/tex]

Let {h*,k*,p*} be the solution to (1)-(3). Then, the locus of points on the parabola is "the set of all (x,y) in [itex]\mathbb R^2[/itex] that satisfy the standard form equation when {h*,k*,p*} are substituted for {h, k, p} in that equation."
aisha
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#8
Aug26-05, 01:00 AM
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Nope i meant focus not locus p stands for the focus of a parabola i dont understand what to do with all those equations. Can someone tell me how to solve for p?
mantito
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#9
Aug26-05, 05:03 AM
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Quote Quote by aisha
Well yes I think I do know two more points

(0,-50) and (0,50) what do I do to complete my standard form equation?

Do I sub in one of the points into [tex] x^2=4p (y-20) [/tex] and then solve for p?

I dont think i'm doing this right
[tex] x^2=4p (y-20) [/tex]
[tex] 0^2=4p(50-20) [/tex]
[tex] 0=4p(30) [/tex]
[tex] 0=120p [/tex]

now what?? I dont know what to do please help me!
you know points (50,0) and (-50,0). solve equation [tex]x^2=4p (y-20)[/tex] again.
EnumaElish
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#10
Aug26-05, 08:45 AM
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Eq. (1) is how [itex](x-h)^2=4p(y-k)[/itex] looks when x = 0, y = 20.
Eq. (2) is how [itex](x-h)^2=4p(y-k)[/itex] looks when x = -50, y = 0.
Eq. (3) is how [itex](x-h)^2=4p(y-k)[/itex] looks when x = 50, y = 0.
aisha
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#11
Aug27-05, 01:47 AM
P: 587
I'm sorry I see all the equations but they are all the same and I still dont know how to solve for p.
HallsofIvy
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#12
Aug27-05, 11:37 AM
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Quote Quote by aisha
I'm sorry I see all the equations but they are all the same and I still dont know how to solve for p.
No, they are NOT the same! When EnumaElish said
"Eq. (1) is how [itex](x-h)^2= 4p(y-k)[/itex] looks when x = 0, y = 20."

He really expected you to DO that: put x= 0, y= 20 and see what you get. Putting x=0, y= 20, then x= -50, y= 0, then x= 50, y= 0 into
[itex](x-h)^2= 4p(y-k)[/itex] gives you three different equations that you can then solve for h, p, and k. (in fact, you could probably guess h and k!)


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