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anonymous

  • one year ago

HELP PLEASE ON PRECALC, Here's the question, help quickly and correctly and you shall receive a medal! lol: Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, -7).

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  1. anonymous
    • one year ago
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    @phi

  2. anonymous
    • one year ago
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    Hello?

  3. anonymous
    • one year ago
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    Please help, I need to know how to do this, I am working alone online - my teacher won't help and the lesson doesn't make sense. Please help.

  4. anonymous
    • one year ago
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    @phi can you help?

  5. phi
    • one year ago
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    First, plot the focus and vertex (roughly , need not be a work of art) can you do that?

  6. anonymous
    • one year ago
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    sure hold on

  7. anonymous
    • one year ago
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    actually no, can you? not sure where everything goes

  8. phi
    • one year ago
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    vertex is (0,0) and focus is (0,-7) can you plot these points?

  9. anonymous
    • one year ago
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    yep|dw:1437595113760:dw|

  10. phi
    • one year ago
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    and the parabola is a U shape the "curls around" the focus the vertex is the "tip" of the parabola. so it looks like this |dw:1437595121260:dw|

  11. anonymous
    • one year ago
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    ok so how do we get the equation now?

  12. anonymous
    • one year ago
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    uhoh

  13. phi
    • one year ago
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    the standard equation for the simplest parabola (with vertex at 0,0) is y= x^2 this one is upside down, so expect it to be y = - a x^2 to find y, we use the idea that 4 p y = x^2 is the equation of a parabola and "p" is the distance from the vertex to the focus

  14. phi
    • one year ago
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    in this case, the distance from -7 to 0 is 7 (ignoring the sign) 4p is 4*7= 28 so 28y= x^2 y = 1/28 x^2 but it has to be negative because we are upside down so \[ y = - \frac{1}{28} x^2\]

  15. anonymous
    • one year ago
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    so for p i do this: \[\left| -7 - 0 \right| = 7 so p = 7\]

  16. phi
    • one year ago
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    that works

  17. anonymous
    • one year ago
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    oh ok. thanks can you help with one more. same type of problem, different info given

  18. anonymous
    • one year ago
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    Here it is: Find the standard form of the equation of the parabola with a focus at (7, 0) and a directrix at x = -7.

  19. anonymous
    • one year ago
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    i am going to draw it is that ok?

  20. phi
    • one year ago
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    yes, just to get an idea of what is going on.

  21. anonymous
    • one year ago
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    |dw:1437595918471:dw| it looks bad but i think that is the right idea?

  22. anonymous
    • one year ago
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    so what equation do i use for horizontal parabolas

  23. phi
    • one year ago
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    ok, except the vertex is exactly halfway between the directrix and focus

  24. anonymous
    • one year ago
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    oh ok so...

  25. phi
    • one year ago
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    you can just average the x values (-7+7)/2 = 0 the vertex is at (0,0) this one is sideways, so swap x and y in the standard 4py= x^2 to get 4p x = y^2 just like before p is the distance from the vertex to the focus

  26. anonymous
    • one year ago
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    which is 7 right? so p =7

  27. anonymous
    • one year ago
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    4(7)x = y^2?

  28. anonymous
    • one year ago
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    which simplifies to 28x = y^2, but that's not an answer choice. do you want to see them?

  29. phi
    • one year ago
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    ok

  30. anonymous
    • one year ago
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    y = (1/28)x^2 x = (1/28)y^2 -28y = x^2 y^2 = 14x

  31. phi
    • one year ago
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    28x = y^2 divide both sides by 28

  32. anonymous
    • one year ago
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    oh ok awesome! thanks so much!

  33. phi
    • one year ago
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    yw

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