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lilsis76 Group Title

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.

  • 2 years ago
  • 2 years ago

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  1. lilsis76 Group Title
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    Now that I look at the problem, i dont even think I did it right

    • 2 years ago
  2. tkhunny Group Title
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    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?

    • 2 years ago
  3. lilsis76 Group Title
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    let me see if i can work it to a equation with....\[(x-h)^{2} = 4p(y-k) \right?\]

    • 2 years ago
  4. lilsis76 Group Title
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    @tkhunny

    • 2 years ago
  5. tkhunny Group Title
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    Yes, but you already know h = k = 0. No need to mess with those.

    • 2 years ago
  6. lilsis76 Group Title
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    okay um... hold on

    • 2 years ago
  7. lilsis76 Group Title
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    ya, i got nothing :/ I keep getting the zero, like im answering it . i dont know how to find the equation

    • 2 years ago
  8. tkhunny Group Title
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    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.

    • 2 years ago
  9. lilsis76 Group Title
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    so if its 4, i divide by the 4p to get 1

    • 2 years ago
  10. tkhunny Group Title
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    Close. You're backwards. 'p' is the distance that we know. p = 4, then 4p = 16.

    • 2 years ago
  11. lilsis76 Group Title
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    oh okay. so then the equation i put in the 16? this has been a long section and my brain is completelly fried badly

    • 2 years ago
  12. lilsis76 Group Title
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    @tkhunny

    • 2 years ago
  13. tkhunny Group Title
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    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.

    • 2 years ago
  14. lilsis76 Group Title
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    isnt that a different equation to another problem? it looks familar

    • 2 years ago
  15. lilsis76 Group Title
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    how do i get the focus of 4,0?

    • 2 years ago
  16. tkhunny Group Title
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    That's where we started. Remember the cernter of the circle?

    • 2 years ago
  17. lilsis76 Group Title
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    isnt that 0,0? right the center? cuz the vertex is.....is the point where the parabola starts

    • 2 years ago
  18. tkhunny Group Title
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    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)

    • 2 years ago
  19. lilsis76 Group Title
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    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?

    • 2 years ago
  20. tkhunny Group Title
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    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.

    • 2 years ago
  21. lilsis76 Group Title
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    okay, but how do we get the 16? i dont understand how that was found

    • 2 years ago
  22. tkhunny Group Title
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    More Review: The distance from the vertex to the focus is 4. Thus, p = 4 and 4p = 16

    • 2 years ago
  23. lilsis76 Group Title
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    so that means.....you...divide by the 4 to get 4? an that is the focus?

    • 2 years ago
  24. tkhunny Group Title
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    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.

    • 2 years ago
  25. lilsis76 Group Title
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    i will thank you. i will go over these steps.

    • 2 years ago
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