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anonymous

  • one year ago

Find the standard form of the equation of the parabola with a focus at (-8, 0) and a directrix at x = 8.

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  1. triciaal
    • one year ago
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    |dw:1433220174652:dw|

  2. triciaal
    • one year ago
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    @jim_thompson5910 please verify

  3. jim_thompson5910
    • one year ago
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    The parabola will open to the left side, so it will have a y^2 term (instead of an x^2 term)

  4. triciaal
    • one year ago
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    not use to those so the vertex is (0,0)

  5. jim_thompson5910
    • one year ago
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    |dw:1433217367354:dw|

  6. jim_thompson5910
    • one year ago
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    |dw:1433217380478:dw|

  7. jim_thompson5910
    • one year ago
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    |dw:1433217399104:dw|

  8. jim_thompson5910
    • one year ago
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    |dw:1433217436087:dw|

  9. jim_thompson5910
    • one year ago
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    I'm using the ideas shown here http://www.mathwords.com/f/focus_parabola.htm 4p(x-h) = (y-k)^2 4*8(x-0) = (y-0)^2 32x = y^2 x = (1/32)*y^2 x = -(1/32)*y^2 ... the coefficient is made to be negative so the parabola opens to the left (instead of right) |dw:1433217556903:dw|

  10. triciaal
    • one year ago
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    so is the equation I have wrong ?

  11. jim_thompson5910
    • one year ago
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    anything with x^2, and not y^2, will open upward or downward

  12. triciaal
    • one year ago
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    I actually have the same diagram see above (not as neat )

  13. jim_thompson5910
    • one year ago
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    I see it, but again, you'll have y^2 instead of x^2

  14. jim_thompson5910
    • one year ago
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    http://images.flatworldknowledge.com/reddenint/reddenint-fig08_009.png

  15. triciaal
    • one year ago
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    |dw:1433221628889:dw|

  16. triciaal
    • one year ago
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    you are saying since I have the directrix vertical instead of horizontal then the value from f to v is the y component?

  17. jim_thompson5910
    • one year ago
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    because the directrix is vertical, this means that the parabola will open to the left or right https://dr282zn36sxxg.cloudfront.net/datastreams/f-d%3A7980ac3c31917d6d9238e3fe5ee79be178774387290a21eba7ba82d2%2BIMAGE%2BIMAGE.1

  18. triciaal
    • one year ago
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    got that not the problem what I am missing is how to write the equation

  19. jim_thompson5910
    • one year ago
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    http://www.mathwords.com/f/focus_parabola.htm you use 4p(x-h) = (y-k)^2 since the parabola opens to the left, p < 0 based on that previous image. So p = -8 (h,k) = (0,0) is the vertex

  20. triciaal
    • one year ago
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    @jim_thompson5910 thank you so much !!! I will look at the links guess better try to learn that version of the equation thank you

  21. anonymous
    • one year ago
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    So what would the finally equation be?

  22. anonymous
    • one year ago
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    *final

  23. anonymous
    • one year ago
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    @jim_thompson5910

  24. triciaal
    • one year ago
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    follow what Jim had above. I am used to the other format and in addition I had my x and y switched since it opened left, sorry

  25. jim_thompson5910
    • one year ago
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    It depends on what format they want it in, but you could go with something in the form x = ay^2 + by + c

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