Question

Derive the equation of the parabola with a focus at (0, −4) and a directrix of y = 4.

Answer 1

I miss this stuff :D
Got any idea what focus and directrix mean?

Answer 2

To be honest, not a clue.

Answer 3

If you can tell me the equation of this parabola, so that we can start on the calculus part of your problem.

Answer 4

That's the whole question.

Answer 5

I can see what you are asked, but I tried to nicely ask you to find the equation of the parabola first.

Answer 6

I'm sorry. I don't know how to do that either. This is my weak point in math.

Answer 7

Okay, you know it is a vertical parabola, since it's directrix is "y="

Answer 8

...great. The OP left.
@SolomonZelman You basically asked the OP to do the question himself :/

Answer 9

@Supreme_Kurt maybe, if you don't know how to help the user, you can try not to interrupt?
That would be great.

Answer 10

Okay, mdf,
So, your focus is (0,-4), and by "0" the x coordinate of the minimum point is given.
Now, the directrix, gives it's y-value, and in this case it is y=4.

So you know that the minimum point is (0,4)

\(\large\color{black}{ (x-h)^2=4p(y-k) }\)
would be essentially giving the equation of the parabola.

You know that the focus is 8 units away from the vertex.

Answer 11

\(\large\color{black}{ (x-0)^2=4(8)(y-4) }\)
See how I am plugging it in?

Answer 12

\(\large\color{black}{ x^2=32y-128 }\)
\(\large\color{black}{ 32y=- x^2-128 }\)
\(\large\color{black}{ y=-\frac{1}{32}x^2-4 }\)

Answer 13

To derive it, you shall use just the power rule.

Answer 14

Good thing the OP has left. Your answer is wrong. Care to try again?

Answer 15

yes, I would want to remember how to do it correctly-:(

Answer 16

I'll try on my paper.

Answer 17

First, why don't you apologize to your former tutor? :P

Answer 18

Anyway, you're right in essence. the equation of a vertical parabola is
\[\Large (x-h)^2 = 4p(y-k)\] with (h,k) being the vertex.

Answer 19

First problem: Wrong vertex.
care to fix that?
What's the correct vertex?

Answer 20

thew vertex is between.

Answer 21

(0,0)
so I would just plug in 4, and it would be \(\large\color{black}{(x-0)^2=4(4)(y-0) }\)
\(\large\color{black}{x^2=16y }\)

\(\large\color{black}{\frac{1}{16}x^2=y }\)

Answer 22

Still no.

Answer 23

Why would you plug in p = 4?

Answer 24

the distance between vertex and focus is 4?

Answer 25

yeah... math was never my thing. I'll watch a video and be back I guess.

Answer 26

Well, you could put it that way.
But I prefer to think of p as the difference between the vertex y-value and the focus y-value (for vertical parabolas).

In that case, p=-4

That way, you can immediately see that it's a downward-facing parabola with a MAXIMUM rather than a minimum, as you stated...

Answer 27

:) it's over -16.

Answer 28

I truely suck

Answer 29

truly

Answer 30

No, you don't :/
You just need practice.
And humility. Never forget humility, ok? ^^

Answer 31

truly

Answer 32

EVeryone sucks in his own way, or at least anyone. But I think I learned again.
I watched that video with -p