Question

The vertex form of the equation of a vertical parabola is given by y= 1/4p(x-h)^2 , where (h, k) is the vertex of the parabola and the absolute value of p is the distance from the vertex to the focus, which is also the distance from the vertex to the directrix. You will use the GeoGebra geometry tool to create a vertical parabola and write the vertex form of its equation. Open GeoGebra, and complete each step below. If you need help, follow these instructions for using GeoGebra.
Part A:

Mark the focus of the parabola you are going to create at F(6, 4). Draw a horizontal line that is 6 units below the focus. This line will be the directrix of your parabola. What is the equation of the line?


Part B:

Construct the line that is perpendicular to the directrix and passes through the focus. This line will be the axis of symmetry of the parabola. What are the coordinates of the point of intersection, A, of the axis of symmetry and the directrix of the parabola?


Part C:

Explain how you can locate the vertex, V, of the parabola with the given focus and directrix. Write the coordinates of the vertex.


Part D:

Which way will the parabola open? Explain.


Part E:

How can you find the value of p? Is the value of p for your parabola positive or negative? Explain.


Part F:

What is the value of p for your parabola?


Part G:

Based on your responses to parts C and E above, write the equation of the parabola in vertex form. Show your work.


Part H:

Construct the parabola using the parabola tool in GeoGebra. Take a screenshot of your work, save it, and insert the image below.


Part i:

Once you have constructed the parabola, use GeoGebra to display its equation. In the space below, rearrange the equation of the parabola shown in GeoGebra, and check whether it matches the equation in the vertex form that you wrote in part G. Show your work.


THANK YOU!!!!!

Answer 1

Answer:

  A. y = -2

  B. A(6, -2)

  C. use the midpoint tool to locate the vertex B(6, 1) halfway between F and A

  D. up

  E. divide 6 by 2; positive

  F. 3

  G. y = (1/12)(x -6)² +1

  H. see attached

Step-by-step explanation:

A. The horizontal line 6 units below a point with y-coordinate 4 will be at ...

  y = 4 -6

  y = -2

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B. The point of intersection will have the same x-coordinate as the focus (6) and the same y-coordinate as the line (-2). That point is A(6, -2).

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C. As with every other point on the parabola, the vertex is the same distance from the focus as it is from the directrix. It will be the midpoint of segment FA. The vertex is B(6, 1).

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D. The focus is "inside" the parabola. The focus is above the directrix, so the parabola opens upward, away from the directrix.

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E, F. We're told to locate the directrix 6 units below the focus. The value of p is half that distance, 3 units. It can also be found by using GeoGebra to measure the length of segment FB.

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G. Using the vertex form equation with p=3, (h, k) = (6, 1), the equation of the parabola is ...

  $y=\dfrac{1}{4p}(x-h)^2+k=\dfrac{1}{4\cdot 3}(x -6)^2+1\\\\\boxed{y=\dfrac{1}{12}(x-6)^2+1}$

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H. see the attachment

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I. Multiplying the above equation by 12 and eliminating parentheses, we have ...

  12y = x^2 -12x +36 +12

  -48 = x^2 -12x -12y . . . . . . . subtract 48+12y

This is the same equation as shown by GeoGebra.