Question

Find the standard form of the equation of the parabola with a focus at (0, 8) and a directrix at y = -8

Answer 1

would the answer be y = 1/8x^2?

Answer 2

The distance between the focus and the vertex, and the distance between the vertex and the directrix, is a constant \(p\). \[4p(y-k) = (x-h)^2\]where the vertex is at \((h,k)\)

Answer 3

So 8 right?

Answer 4

You tell me! But remember that the definition of a parabola is all the points equidistance from the focus and the directrix. Here's a graph of your parabola, and I've marked in two lines at x=8, y=8. The directrix is the purple line across the bottom. Does the point where the two lines intersect look like it is an equal distance from (0,8)?

Answer 5

yes

Answer 6

NO! From (0,8) to (8,8) is 8 units. From (8,-8) to (8,8) is 16 units.

Answer 7

Why don't you show me how you came up with your formula, so we can figure out the mistake?

Answer 8

Oh, I misread your comment. When you said " Does the point where the two lines intersect look equal" I looked at (0,8) to (8,8) and (0,8) to (8,-8)

Answer 9

Oh, yeah, then it would look equal, I agree :-)

Answer 10

But do you agree that the point over at (8,8) is not equally far from (0,8) and the nearest approach of the directrix at (8,-8)?

Answer 11

agreed

Answer 12

That means that point is not on the parabola we want.

Answer 13

Do you know the vertex form equation for a parabola?

Answer 14

y= a(x-h)^2 + k?

Answer 15

That's one version of it, yes. We will find it more convenient to rearrange that a bit:
\[y-k = a(x-h)^2\]\[\frac{1}{a}(y-k) = (x-h)^2\]
We can still "read out" the vertex from that, agreed?

Answer 16

yeah

Answer 17

Okay, I'm going to change it to \[4p(y-k) = (x-h)^2\]\[4p=\frac{1}{a}\]\[p=\frac{1}{4a}\]

\(p\) is the distance between the focus and the vertex, and the distance between the vertex and the directrix.

We know those distances here, don't we?

Answer 18

1/32?

Answer 19

what's 1/32?

Answer 20

p =1/4(8) = 1/32
So the answer would be y= 1/32 x^2?

Answer 21

or am I jumping the gun?

Answer 22

yes, \[y = \frac{1}{32}x^2\]is the correct answer.

Answer 23

thank you

Answer 24

In this graph, the blue line is your original answer, the purple is the new, improved answer, the olive line is the directrix, and the two black lines I added intersect at (16,8), which you can this time is equally far from the focus (0,8) and the directrix y = -8

Answer 25

you got the right answer but did the math wrong :-) p = 8 here. p is the distance between focus and vertex: (0,8) to (0,0) = 8 and also the distance between vertex and directrix: (0,0) to y = -8 = 8

Answer 26

If you have your equation in the \[4p(y-k)=(x-h)^2\]variant of vertex form, the vertex is at \((h,k)\), the focus is at \((h,k+p)\) and the directrix is at \(y = k-p\)

Answer 27

but either way I will still end up with y=1/32 x^2 right?

Answer 28

Yes, \[y=\frac{1}{32}x^2\] is the correct answer here. But \(p\ne \frac{1}{32}\) as you wrote above. Perhaps you meant that \(p=8\) and just were a bit sloppy in how you wrote your work.