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

## Answers and Replies

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

Homework Helper
aisha said:
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?

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!

Last edited:
Homework Helper
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? plzzzz help!!

Homework Helper
aisha said:
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."

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?

mantito
aisha said:
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.

Homework Helper
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 dont know how to solve for p.

"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!)