Question

NEED HELP WITH Investigating Quadratics

Answer 1

im here

Answer 2

are you good with math

Answer 3

ummm yes

Answer 4

are you good with quadratics

Answer 5

Quadratic equations???

Answer 6

I love quadratics! I can help!

Answer 7

i dont know what r u working with

Answer 8

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

Answer 9

that would be 0 right

Answer 10

Give me a moment, I need to refresh mah brain with parabolas :)

Answer 11

lol no its not 0

Answer 12

Are you given options?

Answer 13

Using the parameters of a parabola

https://en.wikipedia.org/wiki/Parabola

We have that p=-4

Since the directrix is 4 and the focus is at (0,-4), we have a parabola that points down:

Answer 14

We have
$$
x^2=4py\\
x^2=-16y
$$
http://www.wolframalpha.com/input/?i=x%5E2%3D-16y

Does this make sense?

Answer 15

wait so what

Answer 16

Would you like me to show you an easier way?

Answer 17

well i guess u dont need my help

Answer 18

yes pls

Answer 19

Okay, I just watched the video on this from Kahn, since it's been well over a year since I did it. The equation that I wrote out after watching the video was
\[\sqrt{(x-0)^2+(y-(-4))^2}=\sqrt{(y-4)^2}\]He does the square and square root to make sure it's always positive. After that, it's just down to simplifying and getting it into the y-_=_(x-_)^2 formula

Answer 20

So, after writing that equation out that I found, we'd need to take the square of both sides to get rid of those nasty sqrts. We also need to expand the (y+_)^2s \[(x)^2+y^2+8y+16=y^2-8y+16\]
Then, subtract the \[y^2+8y+16\]from both sides to get
\[x^2=-16y\]And we need to get y by itself. Is this making sense so far?

Answer 21

-1/16 * -16 leaves us with a coefficient of one, so that's what we will multiply \(x^2\) by.
\[y-0=-\frac{1}{16}(x)^2\]or\[y=-\frac{1}{16}(x)^2\]

Answer 22

wait so the first one is the formula ?

Answer 23

That's just finding the distance from any point on the parabola from the focus if that makes sense.

Answer 24

https://www.khanacademy.org/math/algebra2/conics_precalc/parabolas_precalc/v/using-the-focus-and-directrix-to-find-the-equation-of-a-parabola

Answer 25

oh okay thanks

Answer 26

No problem!

Answer 27

Derive the equation of the parabola with a focus at (-5, 5) and a directrix of y = -1

Answer 28

Do the same thing as last time. So we get \[\sqrt{(x-(-5))^2+(y-5)^2}=\sqrt{(y-(-1))^2}\]

Answer 29

this is spaming me and its making me mad

Answer 30

Well I'm sorry I'm helping a person out.

Answer 31

ill open a new question

Answer 32

thx