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anonymous

  • one year ago

Find the standard form of the equation of the parabola with a focus at (5, 0) and a directrix at x = -5.

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  1. anonymous
    • one year ago
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    Center is (0,0)

  2. anonymous
    • one year ago
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    @campbell_st

  3. campbell_st
    • one year ago
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    have you sketched the focus and directrix to determine which way the parabola opens...?

  4. anonymous
    • one year ago
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    It opens right

  5. campbell_st
    • one year ago
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    there are 2 ways to do this question... 1. use the distance formula... knowing the distance from the focus to a point on the parabola is equal to the distance from the point to the directirx. 2. find the focal length and use a standard form of the parabola.

  6. campbell_st
    • one year ago
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    great so the standard from I use for this parabola is \[(y - k)^2 = 4a(x - h)\] where (h, k) is the vertex and a is the focal length. any ideas on the focal length..?

  7. anonymous
    • one year ago
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    I have no idea what that is.

  8. anonymous
    • one year ago
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    wait. it's 5

  9. anonymous
    • one year ago
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    I just remembered what that was

  10. campbell_st
    • one year ago
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    ok... so it seems you have only been taught about selecting a point on the parabola.... |dw:1436392195951:dw| do by definition, the distance from the focus to P is equal to the distance from P to Q I'd expect that's the way you have been taught to find the locus of the parabola.

  11. campbell_st
    • one year ago
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    |dw:1436392369178:dw| so which method do you want to use..?

  12. campbell_st
    • one year ago
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    the line of symmetry for the parabola is y = 0 the vertex is halfway between the directrix and focus.. so any ideas of the location of the vertex and the focal length..?

  13. anonymous
    • one year ago
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    I dont care. Whichever you are better with or think is easier.

  14. anonymous
    • one year ago
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    Vertex is (0,0) and focal length is 5

  15. campbell_st
    • one year ago
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    ok... so how far between the directrix and focus?

  16. anonymous
    • one year ago
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    10

  17. campbell_st
    • one year ago
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    great so 2a = 10 so a = 5 is that ok...?

  18. anonymous
    • one year ago
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    yup

  19. campbell_st
    • one year ago
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    the vertex in this question is to the left of the focus... and it is the focal length to the left on the line of symmetry any ideas in where the vertex is..?

  20. anonymous
    • one year ago
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    0,0?

  21. campbell_st
    • one year ago
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    great so you know h = 0, k =0 and a = 5 plug them into the standard form \[(y - k)^2 = 4a(x - h)\] (h, k) is the vertex and a is the focal length

  22. anonymous
    • one year ago
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    one sec

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