# Loci Conics Parabola eqn in standard form

1. Aug 24, 2005

### 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 im not sure what the focus is or how to determine it plz help me out

2. Aug 24, 2005

### Hurkyl

Staff Emeritus
Do you know of any other points on your parabola?

3. Aug 25, 2005

### HallsofIvy

Staff Emeritus
What are the coordinates of the ends of the bridge?

4. Aug 25, 2005

### 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 dont think i'm doing this right
$$x^2=4p (y-20)$$
$$0^2=4p(50-20)$$
$$0=4p(30)$$
$$0=120p$$

Last edited: Aug 25, 2005
5. Aug 25, 2005

### HallsofIvy

Staff Emeritus
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. Aug 25, 2005

### aisha

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

7. Aug 25, 2005

### EnumaElish

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."

8. Aug 26, 2005

### aisha

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. Aug 26, 2005

### mantito

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

10. Aug 26, 2005

### 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.

11. Aug 27, 2005

### 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.

12. Aug 27, 2005

### HallsofIvy

Staff Emeritus
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!)