Question

what is the focus of the parabola given by the equation y= -1/28 x^2 ?

Answer 1

please explain, thankyouuu!

Answer 2

uhh. not really. this whole lesson confuses me!
I know what a vertex is, I just dont know what it is in this problem

Answer 3

This might help:
http://www.sciencebuddies.org/science-fair-projects/project_ideas/CompSci_img011.gif

Answer 4

but wait how do I find the focus?

Answer 5

http://www.math-prof.com/Alg2/Alg2_Ch_40.asp

Answer 6

so I try and get the x^2 by itself?

Answer 7

\[ y= -\frac{1}{28}x^2 \]
\[-28y=x^2\]
\[4p=-28\]
\[p=-7\]

Answer 8

Hey! How you did the 3rd step? @satellite73

Answer 9

Nice question @mother123

Answer 10

wait yeah how did you do the third step, Im lost.

Answer 11

When Satellite's near, we don't need to worry! :D

Answer 12

haha so the answer is (-7,0) or (0,-7)
Those are the choices I have for my multiple choice

Answer 13

your goal is to make it look like \(4py=x^2\) so you can find \(p\) and then find the focus

Answer 14

once we have \(-28y=x^2\) we know \(4p=-28\) and so we know \(p=-7\) but we are not done yet

Answer 15

so 4py=x^2 is something like a formula?

Answer 16

yes old school formula or more generally \(4p(y-k)=(x-h)^2\)

Answer 17

now to actually find the focus, we draw a picture

Answer 18

since the parabola has vertex at origin and opens down, we see the focus is \((0,-7)\)

Answer 19

so then the focus is (0,-7)?

Answer 20

Perfect!!!

Last question: Now what's the directix?
@satellite73

Answer 21

ohh I get it a little more! wait I have another one, can you tell me If Im doing it right?

Answer 22

I will try that too!

Answer 23

evidently since the directrix is a line same distance from the vertex as the focus, it should be \(y=7\)

Answer 24

x=3(y-1)^2+2

Answer 25

do I divide everything by 3?

Answer 26

What's the exact meaning of directix?

Answer 27

Why is it there?

Answer 28

@mother123 , you should have x^2 as the subject if in not wrong?

Answer 29

im*

Answer 30

one definition of the parabola is all points whose distance from a fixed point (focus) and a line (directrix) are equal

Answer 31

Satellite is the superman here:
http://femour.com/wp-content/uploads/2012/07/tumblr_m7h2z1Basv1qiekquo1_500.jpg

Answer 32

idk. there is no x^2 in the problem

Answer 33

\[x=3(y-1)^2+2\]
\[x-2=3(y-1)^2\]
\[\frac{1}{3}(x-2)=(y-1)^2\]

Answer 34

What's the original question?

Answer 35

don't lose track of the goal
you want to make it look like
\(4p(x-h)=(y-k)^2\) or
\[4p(y-k)=(x-h)^2\]

Answer 36

parabola that opens up or down, with vertex at \((h,k)\) will look like
\[4p(y-k)=(x-h)^2\] and parabola that opens left or right with vertex \((h,k)\) will look like
\[4p(x-h)=(y-k)^2\]

Answer 37

how do you get 1/3 to be a 4?

Answer 38

i see..

Answer 39

the (h\) always goes with the \(x\) and the \(k\) always goes with the \(y\)
make it iin to this form to get what you want

Answer 40

@mother123 you do not make \(\frac{1}{3}=4\) (which would be a miracle) you make \(4p=\frac{1}{3}\) and solve for \(p\)

Answer 41

so p = 1/12

Answer 42

This is the hardest thing I've ever done. Im so confused.

Answer 43

yes

Answer 44

idk if this is right?