parabolas

• Apr 1st 2008, 05:42 PM
danielle1
parabolas
how do you find the focus and the directrix line if all you have is the equation in parabolic form?
• Apr 1st 2008, 05:45 PM
Jhevon
Quote:

Originally Posted by danielle1
how do you find the focus and the directrix line if all you have is the equation in parabolic form?

see post #2 here
• Apr 1st 2008, 05:47 PM
danielle1
woah i have no idea where that p came from.
• Apr 1st 2008, 05:49 PM
Jhevon
Quote:

Originally Posted by danielle1
woah i have no idea where that p came from.

it is just a general form. all the letters but x and y represent constants. your objective is to try and get your parabola into that form. then you can figure out what p and the other letters are
• Apr 1st 2008, 05:53 PM
danielle1
ughh i wish my teacher would do his job. i have no idea what you mean. like if i have y=(x-2)+3, i know how to get the vertex but not the focus or the directrix
• Apr 1st 2008, 05:54 PM
Jhevon
Quote:

Originally Posted by danielle1
ughh i wish my teacher would do his job. i have no idea what you mean. like if i have y=(x-2)+3, i know how to get the vertex but not the focus or the directrix

that's not a parabola
• Apr 1st 2008, 05:58 PM
danielle1
wait.... what??? why not?
• Apr 1st 2008, 06:06 PM
Jhevon
Quote:

Originally Posted by danielle1
wait.... what??? why not?

the x is not squared. i suppose you meant to type $y = (x - 2)^2 + 3$

now, you should know that if a parabola is in the form $y = a(x - h)^2 + k$, the vertex is given by $(h,k)$...oh, you do know that. ok, moving on.

now we want to get this parabola to look like the form i used in the post i directed you to.

the form must look like $(x - h)^2 = 4p(y - k)$

in this form, the focus is: $(h, k + p)$

and the directrix is: $y = k - p$

now, $y = (x - 2)^2 + 3$

$\Rightarrow (x - 2)^2 = y - 3$

the coefficient of y is 1, so where would the 4 come from? well, we can notice that $1 = \frac 44 = 4 \frac 14$

thus, we have $(x - 2)^2 = 4 \frac 14y - 3 = 4 \frac 14(y - 3)$

so, we have $(x - 2)^2 = 4 \cdot \frac 14(y - 3)$ ....look familiar?
• Apr 1st 2008, 06:10 PM
danielle1
omg i think i actually get it now. thank you so much!
• Apr 1st 2008, 06:11 PM
Jhevon
Quote:

Originally Posted by danielle1
omg i think i actually get it now. thank you so much!

ok, what is the answer then?

(i will have to leave for a while, maybe a half hour or so, so don't get upset with me if i don't verify your answer right away)
• Apr 1st 2008, 06:23 PM
danielle1
hmm.. what do i do with the p again?
• Apr 1st 2008, 06:41 PM
Jhevon
Quote:

Originally Posted by danielle1
hmm.. what do i do with the p again?

you just said you got it.

look back at my post. what did i say the focus was? what did i say te directrix was?
• Apr 1st 2008, 06:43 PM
danielle1
k+p, but i don't know what to plug in for p
• Apr 1st 2008, 06:55 PM
Jhevon
Quote:

Originally Posted by danielle1
k+p, but i don't know what to plug in for p

it seems you are not reading through my posts. first of all, i never said anything was k + p. neither the focus nor the directrix is k + p. the focus is a coordinate and the directrix is a horizontal line.

i am sorry, but i really do not see the problem you are having. i went through the algebraic manipulations and tried to make it plain. but let's try again.

$(x - {\color{red}h})^2 = 4{\color{blue}p}(y - {\color{green}k})$ ...........this is the form we want
....... $\downarrow$......... $\downarrow$....... $\downarrow$
$(x - {\color{red}2})^2 = 4 {\color{blue}\frac 14}(y - {\color{green}3})$ ............this is the form we got our equation in

now, the focus is: $(h, k + p)$

and the directrix is: $y = k - p$

can you tell me what h, p and k are?
• Apr 1st 2008, 06:58 PM
danielle1
i meant k+p as the y coordinate. and i didn't realize that p is 1/4.... here i'll try again now