SOLUTION: using the formulas for parabolas, graph and write the equations of parabolas with the following properties. Graph an extra point other than a vertex. The Parabola with focus (0, 5)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: using the formulas for parabolas, graph and write the equations of parabolas with the following properties. Graph an extra point other than a vertex. The Parabola with focus (0, 5)      Log On


   

Question 40448: using the formulas for parabolas, graph and write the equations of parabolas with the following properties. Graph an extra point other than a vertex. The Parabola with focus (0, 5) and directrix y= -5.
Found 2 solutions by Nate, stanbon:
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
From the information we can tell that the vertex is at the origin.
P = 1/(4a) when P is the distance from the vertex to the focus and from the vertex to the directrix
5 = 1/(4a)
20a = 1
a = (1/20)
Use:
y+=+a%28x+-+h%29%5E2+%2B+k
y+=+%281%2F20%29%28x%29%5E2
For another point, pick x+=+1 ....
y+=+%281%2F20%29%281%29%5E2
y+=+%281%2F20%29
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 0.05x%5E2%2B0x%2B0+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%280%29%5E2-4%2A0.05%2A0=0.

Discriminant d=0 is zero! That means that there is only one solution: x+=+%28-%280%29%29%2F2%5C0.05.
Expression can be factored: 0.05x%5E2%2B0x%2B0+=+0.05%28x-0%29%2A%28x-0%29

Again, the answer is: 0, 0. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+0.05%2Ax%5E2%2B0%2Ax%2B0+%29

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
using the formulas for parabolas, graph and write the equations of parabolas with the following properties. Graph an extra point other than a vertex. The Parabola with focus (0, 5) and directrix y= -5.
Draw the picture. Do you see that the parabola must be opening upward.
Therefore the general form is x^2=4py
P is half the distance from the directrix to the focus or p=5
EQUATION:
x^2=4(5)y
y=(1/20)x^2
An extra point would be (1,1/20)
Cheers,
Stan H.