# Loci Conics Parabola eqn in standard form

by aisha
Tags: conics, form, loci, parabola, standard
 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 $$(x-h)^2=4p(y-k)$$ 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
 PF Patron Sci Advisor Emeritus P: 16,094 Do you know of any other points on your parabola?
PF Patron
Thanks
Emeritus
P: 38,416
 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 $$(x-h)^2=4p(y-k)$$ 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
What are the coordinates of the ends of the bridge?

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

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

 PF Patron Sci Advisor Thanks Emeritus P: 38,416 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?
 P: 587 that's what I have done in the last post....I'm still not sure what to do? plzzzz help!!
HW Helper
P: 2,482
 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 $$(x-h)^2=4p(y-k)$$ 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."
 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?
P: 13
 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 $$x^2=4p (y-20)$$ and then solve for p? I dont think i'm doing this right $$x^2=4p (y-20)$$ $$0^2=4p(50-20)$$ $$0=4p(30)$$ $$0=120p$$ now what?? I dont know what to do please help me!
you know points (50,0) and (-50,0). solve equation $$x^2=4p (y-20)$$ again.
 HW Helper Sci Advisor P: 2,482 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.
 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.
PF Patron
"Eq. (1) is how $(x-h)^2= 4p(y-k)$ looks when x = 0, y = 20."
$(x-h)^2= 4p(y-k)$ gives you three different equations that you can then solve for h, p, and k. (in fact, you could probably guess h and k!)