
#1
Aug2405, 08:21 PM

P: 587

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] (xh)^2=4p(yk) [/tex] i know the vertex is (0,20) so the equation should look like [tex] x^2=4p (y20) [/tex] so far but im not sure what the focus is or how to determine it plz help me out 



#2
Aug2405, 08:24 PM

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PF Gold
P: 16,101

Do you know of any other points on your parabola?




#3
Aug2505, 09:18 AM

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PF Gold
P: 38,902





#4
Aug2505, 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 (y20) [/tex] and then solve for p? I dont think i'm doing this right [tex] x^2=4p (y20) [/tex] [tex] 0^2=4p(5020) [/tex] [tex] 0=4p(30) [/tex] [tex] 0=120p [/tex] now what?? I dont know what to do please help me! 



#5
Aug2505, 07:07 PM

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PF Gold
P: 38,902

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?




#6
Aug2505, 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!!




#7
Aug2505, 11:20 PM

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P: 2,483

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



#8
Aug2605, 01:00 AM

P: 587

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?




#9
Aug2605, 05:03 AM

P: 13





#10
Aug2605, 08:45 AM

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P: 2,483

Eq. (1) is how [itex](xh)^2=4p(yk)[/itex] looks when x = 0, y = 20.
Eq. (2) is how [itex](xh)^2=4p(yk)[/itex] looks when x = 50, y = 0. Eq. (3) is how [itex](xh)^2=4p(yk)[/itex] looks when x = 50, y = 0. 



#11
Aug2705, 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.




#12
Aug2705, 11:37 AM

Math
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PF Gold
P: 38,902

"Eq. (1) is how [itex](xh)^2= 4p(yk)[/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](xh)^2= 4p(yk)[/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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