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lilsis76

  • 2 years ago

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.

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  1. lilsis76
    • 2 years ago
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    Now that I look at the problem, i dont even think I did it right

  2. tkhunny
    • 2 years ago
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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?

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

  4. lilsis76
    • 2 years ago
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    @tkhunny

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

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

  7. lilsis76
    • 2 years ago
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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

  8. tkhunny
    • 2 years ago
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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.

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

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

  11. lilsis76
    • 2 years ago
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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

  12. lilsis76
    • 2 years ago
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    @tkhunny

  13. tkhunny
    • 2 years ago
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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.

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

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

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

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

  18. tkhunny
    • 2 years ago
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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)

  19. lilsis76
    • 2 years ago
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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?

  20. tkhunny
    • 2 years ago
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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.

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

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

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

  24. tkhunny
    • 2 years ago
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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.

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

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