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

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- Apr 1st 2008, 05:42 PMdanielle1parabolas
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 PMJhevon
see post #2 here

- Apr 1st 2008, 05:47 PMdanielle1
woah i have no idea where that p came from.

- Apr 1st 2008, 05:49 PMJhevon
- Apr 1st 2008, 05:53 PMdanielle1
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 PMJhevon
- Apr 1st 2008, 05:58 PMdanielle1
wait.... what??? why not?

- Apr 1st 2008, 06:06 PMJhevon
the x is not squared. i suppose you meant to type $\displaystyle y = (x - 2)^2 + 3$

now, you should know that if a parabola is in the form $\displaystyle y = a(x - h)^2 + k$, the vertex is given by $\displaystyle (h,k)$...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 $\displaystyle (x - h)^2 = 4p(y - k)$

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

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

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

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

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

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

so, we have $\displaystyle (x - 2)^2 = 4 \cdot \frac 14(y - 3)$ ....look familiar? - Apr 1st 2008, 06:10 PMdanielle1
omg i think i actually get it now. thank you so much!

- Apr 1st 2008, 06:11 PMJhevon
- Apr 1st 2008, 06:23 PMdanielle1
hmm.. what do i do with the p again?

- Apr 1st 2008, 06:41 PMJhevon
- Apr 1st 2008, 06:43 PMdanielle1
k+p, but i don't know what to plug in for p

- Apr 1st 2008, 06:55 PMJhevon
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.

$\displaystyle (x - {\color{red}h})^2 = 4{\color{blue}p}(y - {\color{green}k})$ ...........this is the form we want

.......$\displaystyle \downarrow$.........$\displaystyle \downarrow$.......$\displaystyle \downarrow$

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

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

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

can you tell me what h, p and k are? - Apr 1st 2008, 06:58 PMdanielle1
i meant k+p as the y coordinate. and i didn't realize that p is 1/4.... here i'll try again now