## anonymous one year ago Find the standard form of the equation of the parabola with a focus at (0, -9) and a directrix y = 9.

1. anonymous

@ganeshie8

2. anonymous

@UsukiDoll

3. anonymous

@Astrophysics

4. anonymous

i know how to do it when they give me x but they gave my y instead

5. Astrophysics

|dw:1435907166360:dw| ok first lets just plug what we have, then we will see what formula to use.

6. Astrophysics

The graph is sort of bad, but we can fix that after. We will have to use the equation $(x-h)^2=4p(y-k)$ look familiar, so far so good?

7. anonymous

yea

8. anonymous

we can plug in h k and y

9. anonymous

but not x and p

10. Astrophysics

Ok so lets just write out what we have so far, h = 0 k = -9 Any idea what p might be? This is a bit tricky.

11. anonymous

x = h - p

12. Astrophysics

Well, let me ask you a question, what way is the parabola, is it opening up, down, left, right what's going on?

13. Astrophysics

Also, we can figure out the vertex right away, as it's the same distance from vertex to directrix and vertex to focus. So it's the middle point between the focus and directrix.

14. Astrophysics

|dw:1435908052023:dw| right?

15. anonymous

so the directrix would be 0?

16. Astrophysics

No no, we're given the directrix, it's y = 9, but the vertex is (0,0) as it's the middle point between the directrix and the focus, where we were given the focus (0,-9)

17. anonymous

ohhhh ok

18. Astrophysics

Ok cool, so our graph looks something like this |dw:1435908205110:dw| now we can write the equation

19. Astrophysics

Because it cannot cross the directrix, so if it was the other way, this would happen |dw:1435908273394:dw| and it can't cross the directrix!

20. Astrophysics

|dw:1435908337698:dw|

21. anonymous

so p would be 10?

22. Astrophysics

Nope, not 10, but 9, sorry about the picture, I just put 10 there to give an idea of the scale.

23. anonymous

oh ok yea i see the 9 now

24. Astrophysics

Alright cool, now just plug in all the value and standard form is in the following: $x=ay^2+by+c$

25. anonymous

but in the answer choice the closest thing to 9 for p is -9, would -9 be right?

26. Astrophysics

Yup, -9 is good

27. anonymous

oh wait isnt it 1/4p?

28. anonymous

y^2= 1/4px

29. Astrophysics

30. anonymous

y = -1/9x^2 y2 = -36x y = -1/36x y2 = -9x

31. anonymous

wouldnt it be c

32. Astrophysics

Ok I see so it wants it in $y = \frac{ 1 }{ 4 p}(x-h)^2+k$ form

33. Astrophysics

That looks good to me

34. anonymous

ok thank you! could you help me with one more but ill make a new post so i can give u another medal

35. Astrophysics

That should be x^2 though

36. Astrophysics

y=-1/36x^2

37. anonymous

yea its x^2 i forgot to put the ^2

38. Astrophysics

Ok :)

39. Astrophysics

$y = -\frac{ 1 }{ 36 }x^2$ so just in case, that is your final answer

40. anonymous

ok thank you