parabola: find equation, given vertex and focus

NEHA

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Oct 27, 2006
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If a parabola's vertex is at (4, -2) and its focus at (4,4), write the equation of this parabola, its directrix, and its axis of symmetry. Graph your parabola with all appropriate parts labeled.

I did:

. . .vertex (h, k) = (4, -2)
. . .focus (h, k + p) = (4, 4), so k + p = 2

. . .y - k = 1/4p (x - h)^2 > equation
. . .y + 2 = 1/4(4) (x - 4)^2 > substituted
. . .y = 1/16 (x - 4)^2 - 2 > solved
. . .p < 0
. . .y = -1 5/6
. . .x = 4

ok now graph it ...how?
 
NEHA said:
If a parabola's vertex is at (4, -2) and its focus at (4,4), write the equation of this parabola, its directrix, and its axis of symmetry. Graph your parabola with all appropriate parts labeled.

I did:

. . .vertex (h, k) = (4, -2)
. . .focus (h, k + p) = (4, 4), so k + p = 2

. . .y - k = 1/4p (x - h)^2 > equation
. . .y + 2 = 1/4(4) (x - 4)^2 > substituted
. . .y = 1/16 (x - 4)^2 - 2 > solved
. . .p < 0
. . .y = -1 5/6
. . .x = 4

ok now graph it ...how?

Hmmm...."p" is the distance from the vertex to the focus. If the vertex is at (4, -2) and the focus is at (4, 4), p = 4 - (-2), or 6. Since the vertex is BELOW, the focus, p is positive....which tells you that the graph is going to open UP.


So, the equation should be

y - (-2) = [1 / (4*6)] (x - 4)<SUP>2</SUP>
y + 2 = (1/24) * (x - 4)<SUP>2</SUP>

or,
y = (1/24) * (x - 4)<SUP>2</SUP> - 2

To draw the graph, you know the vertex; plot that point. You know that the axis of symmetry is the line x = 4. Pick a couple of points on one side of the axis of symmetry (like x = 5, and x = 6). Find y for each of those and plot those points, then plot the "mirror image points" on the other side of the axis of symmetry, That should give you enough information to draw a reasonably good sketch of the graph.

I hope this helps you.
 
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