Question

Can you tell me how to write an equation of a parabola with the given focus and directrix? @mathmale

Answer 1

Focus: (0, 3)
Directrix: y = -1

Answer 2

When you're discussing parabolas and the focus is mentioned, you need to use one of the two forms following:

\[4py=x^2~or~4px=y^2.\]This is for the case where your VERTEX is at the ORIGIN.
Looking at the problem given, can you determine whether your parabola opens up or down or to the right or to the left? Hint: sketch y=-1 and (0,3).

Answer 3

Um up?

Answer 4

Yes. Hope you'll sketch it.
Now, because your graph does open up, we decide it's a "vertical parabola". As before, if the vertex is at the origin, the equation is 4py=x^2.

If the vertex is NOT at the origin, the equation is 4p(y-k)=(x-h)^2, where (x,h) is the vertex.

I know this is probably a bit confusing, Please take a moment to graph the line yu=-1 and the point (the focus) (0,3).
Where do you think the vertex is located?

Answer 5

Hint: it's on the y-axis, because the focus is at (0,3), and the line y=-1 and the y-axis are perpendicular.

Answer 6

(0, 1) ?? I don't really know.. I think I did something wrong cause I have two dots in a straight line with no graph..

Answer 7

What you have is perfect. You have 2 dots in a straight line on the y-axis. Your focus is (0,3). Your vertex is (0,1). And your directrix is the line y=-1.

Notice that the vertex is exactly centered between the focus (0,3) and the directrix y=-1.

Answer 8

Yes thats how I guessed it :D lol

Answer 9

Now please determine how far the vertex is from the focus. How far is the vertex from the directrix?

You're right on.

Let p be the distance from the vertex to the focus. What positive value has it?

Once you have that, substitute everything you know into

4p(y-k)=(x-h)^2, and simplify your result.

Answer 10

Hint: vertex is at (h,k): (0,1) (which you found yourself).

Answer 11

P = (2, 0) ?

4(2)(-1-1) = (0 - 0)^2 ?

Answer 12

Actually, p is just 2. It's a distance, not a point. And it's positive.

So, our 4p(y-k)=(x-h)^2 becomes what? Again: vertex is (0,1); p is 2.

Answer 13

Do not substitute values for x and y, because yu want an equation for the parabola.

Answer 14

Taking 4p(y-k)=(x-h)^2, substitute your knowns.

Answer 15

4(2)(y-1) = (x- 0)^2 ?

Answer 16

Or 4(2)(y-1) = x^2.

simplify this, please.

Answer 17

8(y-1) = x^2

Answer 18

Good! Now divide both sides of the equation by 8, to isolate (y-1).

Answer 19

y = 1/8x^2 + 1 ??

Answer 20

This work is sticky because it could be misinterpreted. To remedy this, please use parentheses. If 8(y-1)=x^2, then y-1 = (x^2) / 8. (Less chance for misinterpretation here.)

Finally, add 1 to both sides. Result?

Answer 21

\[y = \frac{ 1 }{ 8 }x^2 + 1\]

Answer 22

that's great! Please take notes n this procedure for later reference. Best to you, Nicole!

Answer 23

Thank you!