1. ## parabolas

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

2. 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

3. woah i have no idea where that p came from.

4. 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

5. 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

6. 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

7. wait.... what??? why not?

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

9. omg i think i actually get it now. thank you so much!

10. 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)

11. hmm.. what do i do with the p again?

12. 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?

13. k+p, but i don't know what to plug in for p

14. 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.

$\displaystyle (x - {\color{red}h})^2 = 4{\color{blue}p}(y - {\color{green}k})$ ...........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?

15. i meant k+p as the y coordinate. and i didn't realize that p is 1/4.... here i'll try again now