lilsis76
  • lilsis76
find the equation of the parabola with the vertex (0,0), and its focus the center of the circle with the equation: x^2 -8x +y^2 +15 = 0 fixed up: X^2 -8x = -Y^2 - 15/ completing the squar: -8/2 = -4^2 = 16 X^2 -8x +16 = -1(y - 15+16) (x-4)^2 = -1 (y-1) so that means the center is 4,1 right? but how can i figure the rest. this is all i got from my notes.
Mathematics
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SOLVED
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katieb
  • katieb
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lilsis76
  • lilsis76
Now that I look at the problem, i dont even think I did it right
tkhunny
  • tkhunny
No. Nice try, through. This shows you are thinking about it and coming up with a plan. Seriously, good work. The circle is ONLY for finding the Focus of the parabola. You used it for a few other things. You correctly completed the square. This should have given the equation of a circle, \((x-4)^{2} + y^{2} = 1\), which is a circle of radius 1 with center at (4,0). This is ALL we need from the circle. We are done with it. We now have, for the desired parabola: Vertex: (0,0) Focus: (4,0) Can you find the equation of that opening-to-the-right parabola?
lilsis76
  • lilsis76
let me see if i can work it to a equation with....\[(x-h)^{2} = 4p(y-k) \right?\]

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lilsis76
  • lilsis76
@tkhunny
tkhunny
  • tkhunny
Yes, but you already know h = k = 0. No need to mess with those.
lilsis76
  • lilsis76
okay um... hold on
lilsis76
  • lilsis76
ya, i got nothing :/ I keep getting the zero, like im answering it . i dont know how to find the equation
tkhunny
  • tkhunny
The distance from the vertex to the focus is 4. This makes the distance from the vertex to teh directrix also 4. You need to be able to find your '4p' from that information.
lilsis76
  • lilsis76
so if its 4, i divide by the 4p to get 1
tkhunny
  • tkhunny
Close. You're backwards. 'p' is the distance that we know. p = 4, then 4p = 16.
lilsis76
  • lilsis76
oh okay. so then the equation i put in the 16? this has been a long section and my brain is completelly fried badly
lilsis76
  • lilsis76
@tkhunny
tkhunny
  • tkhunny
Part of the task is knowing when you are done. Vertex: (0,0) Focus: (4,0) p (the mysterious parameter): 4 \(y^{2} = 16x\) Done.
lilsis76
  • lilsis76
isnt that a different equation to another problem? it looks familar
lilsis76
  • lilsis76
how do i get the focus of 4,0?
tkhunny
  • tkhunny
That's where we started. Remember the cernter of the circle?
lilsis76
  • lilsis76
isnt that 0,0? right the center? cuz the vertex is.....is the point where the parabola starts
tkhunny
  • tkhunny
Youmust read the probelm statement a couple mroe time. You have become confused. Insta-Review ------ We now have, for the desired parabola: Vertex: (0,0) Focus: (4,0)
lilsis76
  • lilsis76
yes okay well i have this @tkhunny X^2 -8x = -Y^2 - 15/ completing the squar: -8/2 = -4^2 = 16 X^2 -8x +16 = -1(y - 15+16) (x-4)^2 = -1 (y-1) is the focus from the 16?
tkhunny
  • tkhunny
More review: The circle is ONLY for finding the Focus of the parabola. You used it for a few other things. You correctly completed the square. This should have given the equation of a circle, \((x−4)^{2} +y^{2} =1\) , which is a circle of radius 1 with center at (4,0). This is ALL we need from the circle. We are done with it.
lilsis76
  • lilsis76
okay, but how do we get the 16? i dont understand how that was found
tkhunny
  • tkhunny
More Review: The distance from the vertex to the focus is 4. Thus, p = 4 and 4p = 16
lilsis76
  • lilsis76
so that means.....you...divide by the 4 to get 4? an that is the focus?
tkhunny
  • tkhunny
Let's go sequentially, shall we. 1) We know we need a parabola. The posibilities are these: \((x-h)^{2} = 4p(y-k)\) or \((y-k)^{2} = 4p(x-h)\) 2) We know the vertex, (0,0). Now the possibilites are these: \(x^{2} = 4py\) or \(y^{2} = 4px\) -- Just substituting h = 0 and k = 0 and simplifying. 3) Using the circle hint, we determined the focus to be (4,0). Since this is just to the right of the vertex (0,0), we know the parabola opens to the right. Now the posibilities are these: \(y^{2} = 4px\) -- No more "or". It's this kind and not the other. 4) Find 'p'. You need either the distance from the vertex to the focus or the vertex to the directrix. We have the former. p = 4 Thus: \(y^{2} = 4(4)x\) -- Simply substituting the known value. 5) Simplify and we're done. \(y^{2} = 16x\) -- Simply substituting the known value. One thing at a time. Slowly. Systematically. If you start getting confused, don't be afraid to start over and be more careful and more systematic.
lilsis76
  • lilsis76
i will thank you. i will go over these steps.

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