Loci Conics Parabola eqn in standard form

[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

(x-h)^2=4p(y-k)

i know the vertex is (0,20) so the equation should look like x^2=4p (y-20) so far but I am not sure what the focus is or how to determine it please help me out

 

[Hurkyl]

Do you know of any other points on your parabola?

 

[HallsofIvy]

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

(x-h)^2=4p(y-k)

i know the vertex is (0,20) so the equation should look like x^2=4p (y-20) so far but I am not sure what the focus is or how to determine it please help me out

What are the coordinates of the ends of the bridge?

 

[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 x^2=4p (y-20) and then solve for p?

I don't think I'm doing this right
x^2=4p (y-20)
0^2=4p(50-20)
0=4p(30)
0=120p

now what?? I don't know what to do please help me!

 

[HallsofIvy]

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]

that's what I have done in the last post...I'm still not sure what to do? pleasezzz help!

 

[EnumaElish]

but I am 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 (x-h)^2=4p(y-k) 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:

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

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 \mathbb R^2 that satisfy the standard form equation when {h^*,k^*,p^*} are substituted for {h, k, p} in that equation."

 

[aisha]

Nope i meant focus not locus p stands for the focus of a parabola i don't understand what to do with all those equations. Can someone tell me how to solve for p?

 

[mantito]

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 x^2=4p (y-20) and then solve for p?

I don't think I'm doing this right
x^2=4p (y-20)
0^2=4p(50-20)
0=4p(30)
0=120p

now what?? I don't know what to do please help me!

you know points (50,0) and (-50,0). solve equation x^2=4p (y-20) again.

 

[EnumaElish]

Eq. (1) is how (x-h)^2=4p(y-k) looks when x = 0, y = 20.
Eq. (2) is how (x-h)^2=4p(y-k) looks when x = -50, y = 0.
Eq. (3) is how (x-h)^2=4p(y-k) looks when x = 50, y = 0.

 

[aisha]

I'm sorry I see all the equations but they are all the same and I still don't know how to solve for p.

 

[HallsofIvy]

I'm sorry I see all the equations but they are all the same and I still don't know how to solve for p.

No, they are NOT the same! When EnumaElish said
"Eq. (1) is how (x-h)^2= 4p(y-k) 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
(x-h)^2= 4p(y-k) gives you three different equations that you can then solve for h, p, and k. (in fact, you could probably guess h and k!)