Loci Conics Parabola eqn in standard form

Then p is determined. In summary, to find the equation of a parabola in standard form with a domain of {-50 <= x <= 50} and a range of {0 <= y <= 20}, you need to use the coordinates of the vertex and two other points on the parabola (such as the endpoints of the arch under the bridge) to solve for the parameters h, k, and p in the equation (x-h)^2=4p(y-k). This will give you the standard form equation for the parabola, which can then be used to find the focus of the parabola.
  • #1
aisha
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0
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] (x-h)^2=4p(y-k) [/tex]

i know the vertex is (0,20) so the equation should look like [tex] x^2=4p (y-20) [/tex] so far but I am not sure what the focus is or how to determine it please help me out
 
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  • #2
Do you know of any other points on your parabola?
 
  • #3
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

[tex] (x-h)^2=4p(y-k) [/tex]

i know the vertex is (0,20) so the equation should look like [tex] x^2=4p (y-20) [/tex] 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?
 
  • #4
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 (y-20) [/tex] and then solve for p?

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

now what?? I don't know what to do please help me!
 
Last edited:
  • #5
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
that's what I have done in the last post...I'm still not sure what to do? pleasezzz help!
 
  • #7
aisha said:
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 [tex](x-h)^2=4p(y-k)[/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] (0-h)^2=4p(20-k) \ ... \text{ Eq. (1)}[/tex]
[tex] (-50-h)^2=4p(0-k) \ ... \text{ Eq. (2)}[/tex]
[tex] (50-h)^2=4p(0-k) \ ... \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
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?
 
  • #9
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 [tex] x^2=4p (y-20) [/tex] and then solve for p?

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

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

you know points (50,0) and (-50,0). solve equation [tex]x^2=4p (y-20)[/tex] again.
 
  • #10
Eq. (1) is how [itex](x-h)^2=4p(y-k)[/itex] looks when x = 0, y = 20.
Eq. (2) is how [itex](x-h)^2=4p(y-k)[/itex] looks when x = -50, y = 0.
Eq. (3) is how [itex](x-h)^2=4p(y-k)[/itex] looks when x = 50, y = 0.
 
  • #11
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.
 
  • #12
aisha said:
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 [itex](x-h)^2= 4p(y-k)[/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](x-h)^2= 4p(y-k)[/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!)
 

1. What is the standard form of a parabola equation?

The standard form of a parabola equation is y = ax^2 + bx + c, where a, b, and c are constants. This form is also known as vertex form, as it allows us to easily identify the vertex of the parabola.

2. How can I tell if a parabola opens upwards or downwards?

A parabola with a positive coefficient for the x^2 term (a > 0) will open upwards, while a parabola with a negative coefficient (a < 0) will open downwards. This can also be determined by looking at the sign of the coefficient in front of the x term (b). If b is positive, the parabola will open upwards, and if b is negative, it will open downwards.

3. How many solutions does a parabola equation have?

A parabola can have either one, two, or no solutions, depending on the location of the parabola in relation to the x-axis. If the parabola intersects the x-axis at two points, it will have two solutions. If it only touches the x-axis at one point, it will have one solution. And if the parabola does not intersect the x-axis at all, it will have no solutions.

4. How can I find the axis of symmetry for a parabola?

The axis of symmetry for a parabola is a vertical line that passes through the vertex of the parabola. To find this line, we can use the formula x = -b/2a, where a and b are the coefficients in the standard form of the parabola equation. This will give us the x-coordinate of the vertex, and the axis of symmetry will be a vertical line passing through this point.

5. Can the vertex of a parabola be located anywhere on the coordinate plane?

No, the vertex of a parabola must lie on the axis of symmetry, which is a vertical line. Therefore, the x-coordinate of the vertex will always be at a specific point, while the y-coordinate can vary depending on the equation. However, the vertex can be located at any point on the x-axis, depending on the equation.

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