## anonymous one year ago Derive the equation of the parabola with a focus at (0, 1) and a directrix of y = −1.

1. anonymous

@zepdrix

2. zepdrix

eyyy

3. zepdrix

I think I remember you asking this yesterday? :O

4. zepdrix

|dw:1438975490650:dw|

5. zepdrix

Our Vertex is always equidistant from our focus and directrix, ya? So that puts us at 0,0

6. zepdrix

|dw:1438975604620:dw|

7. zepdrix

Umm from here... we can either do it the long way... or use that nice shortcut formula.. im trying to remember what it looks like..

8. zepdrix

Here is that shortcut that I was thinking of: $\large\rm (x - h)^2 = 4p (y - k)$Where our vertex is the point (h,k), and p is the distance between the focus and vertex.

9. anonymous

yes thank you i did ask yesterday but i had trouble at the end and couldnt find a :(

10. anonymous

h and k i got as 0, 0

11. zepdrix

$\large\rm (x-0)^2=4p(y-0)$$\large\rm x^2=4py$Ok good. let's solve for y,$\large\rm y=\frac{1}{4p}x^2$

12. zepdrix

And then, what is the value of p? The distance from the vertex up to the focus?

13. anonymous

wouldnt it be 2? from 1 to -1? or no

14. zepdrix

|dw:1438975963650:dw|No that would be 2p :)

15. anonymous

|dw:1438976050358:dw|is this the focus? at 0,1

16. zepdrix

|dw:1438976120582:dw|yes :) i want to know the distance from here to here from the VERTEX to the FOCUS. not from the DIRECTRIX silly :)

17. anonymous

ohhh 1

18. anonymous

sorry sorry

19. zepdrix

$\large\rm y=\frac{1}{4p}x^2 \qquad\to\qquad y=\frac{1}{4\cdot1}x^2$

20. zepdrix

Yayyy team, we did it! \c:/

21. anonymous

okay i got it now thank you soo much

22. anonymous

:))