anonymous
  • anonymous
Find the standard form of the equation of the parabola with a focus at (0, -9) and a directrix y = 9.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@ganeshie8
anonymous
  • anonymous
@UsukiDoll
anonymous
  • anonymous
@Astrophysics

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More answers

anonymous
  • anonymous
i know how to do it when they give me x but they gave my y instead
Astrophysics
  • Astrophysics
|dw:1435907166360:dw| ok first lets just plug what we have, then we will see what formula to use.
Astrophysics
  • 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?
anonymous
  • anonymous
yea
anonymous
  • anonymous
we can plug in h k and y
anonymous
  • anonymous
but not x and p
Astrophysics
  • 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.
anonymous
  • anonymous
x = h - p
Astrophysics
  • Astrophysics
Well, let me ask you a question, what way is the parabola, is it opening up, down, left, right what's going on?
Astrophysics
  • 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.
Astrophysics
  • Astrophysics
|dw:1435908052023:dw| right?
anonymous
  • anonymous
so the directrix would be 0?
Astrophysics
  • 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)
anonymous
  • anonymous
ohhhh ok
Astrophysics
  • Astrophysics
Ok cool, so our graph looks something like this |dw:1435908205110:dw| now we can write the equation
Astrophysics
  • 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!
Astrophysics
  • Astrophysics
|dw:1435908337698:dw|
anonymous
  • anonymous
so p would be 10?
Astrophysics
  • Astrophysics
Nope, not 10, but 9, sorry about the picture, I just put 10 there to give an idea of the scale.
anonymous
  • anonymous
oh ok yea i see the 9 now
Astrophysics
  • Astrophysics
Alright cool, now just plug in all the value and standard form is in the following: \[x=ay^2+by+c\]
anonymous
  • anonymous
but in the answer choice the closest thing to 9 for p is -9, would -9 be right?
Astrophysics
  • Astrophysics
Yup, -9 is good
anonymous
  • anonymous
oh wait isnt it 1/4p?
anonymous
  • anonymous
y^2= 1/4px
Astrophysics
  • Astrophysics
May I see your answer choices actually
anonymous
  • anonymous
y = -1/9x^2 y2 = -36x y = -1/36x y2 = -9x
anonymous
  • anonymous
wouldnt it be c
Astrophysics
  • Astrophysics
Ok I see so it wants it in \[y = \frac{ 1 }{ 4 p}(x-h)^2+k\] form
Astrophysics
  • Astrophysics
That looks good to me
anonymous
  • anonymous
ok thank you! could you help me with one more but ill make a new post so i can give u another medal
Astrophysics
  • Astrophysics
That should be x^2 though
Astrophysics
  • Astrophysics
y=-1/36x^2
anonymous
  • anonymous
yea its x^2 i forgot to put the ^2
Astrophysics
  • Astrophysics
Ok :)
Astrophysics
  • Astrophysics
\[y = -\frac{ 1 }{ 36 }x^2\] so just in case, that is your final answer
anonymous
  • anonymous
ok thank you

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