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anonymous

  • one year ago

An archway will be constructed over a walkway. A piece of wood will need to be curved to match a parabola. Explain to Maurice how to find the equation of the parabola given the focal point and the directrix (8). d = (square root) (Xv2 - 3)^2 + (Yv2 - 6)^2 d = (square root) (Yv2 - 8)^2 I need to something like -1/-8x^2 + 5/4x + 23/8 = y (example given by school) Its to do an arc. I got to Xv2^2 -6Xv2 + 9 + Yv2^2 -8Yv2 + 16 = ???

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  1. anonymous
    • one year ago
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    You are forgetting the weird exponent that is below the number. I don't know if that effects the equation at all though. What I'm supposed to get is something like \[-\frac{ 1 }{ 8 } x^2 + \frac{ 5 }{ 4 } x + \frac{ 23 }{ 8 } = y\] Which someone how becomes this -0.13x^2 + 1.25x + 2.88 (Not my problem, its their example)

  2. anonymous
    • one year ago
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    Expand it? And give me a second, I'll stitch together some screenshots of what they gave for instructions.

  3. anonymous
    • one year ago
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  4. anonymous
    • one year ago
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    I don't really understand any of this as I was horrible at algebra. I need it like what I said before, "-0.13x^2 + 1.25x + 2.88" so I can put it in stupid geogebra to submit for my project.

  5. anonymous
    • one year ago
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    Because I have to put it into geogebra, or I don't get a grade.

  6. campbell_st
    • one year ago
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    so let's check... the focus is at (3, 6) and directrix is at y = 8

  7. anonymous
    • one year ago
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    yes

  8. campbell_st
    • one year ago
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    great so using the distance formula method pick a point P(x, y) on the parabola distance Focus to P \[d = \sqrt{(x - 3)^2 + (y - 6)^2}\] P to the directrix \[d = \sqrt{x -x)^2 + (y - 8)^2}\] equating and squaring you get \[(x -3)^2 + (y - 6)^2 = (y - 8)^2\] now subtract (y - 6)^2 from both sides \[(x -3)^2 = (y -8)^2 - (y -6)^2 \] does that make sense..?

  9. campbell_st
    • one year ago
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    so simplify the right hand side \[(x-3)^2 = y^2 - 16y + 64 - y^2 + 12y - 36\] which becomes \[(x -3)^2 = -4y + 28\] factor the right hand side \[(x -3)^2 = -4(y - 7)\] does that make sense...?

  10. campbell_st
    • one year ago
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    make y the subject \[-\frac{(x -3)^2}{4} + 7 = y\] you can enter the equation as its written above into Geogebra...

  11. campbell_st
    • one year ago
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    it will then write the equation in expanded form with decimal coefficients...

  12. campbell_st
    • one year ago
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    and I just graphed the parabola in geogebra and it worked perfectly...

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  13. anonymous
    • one year ago
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    how do you graph it in it?

  14. campbell_st
    • one year ago
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    just enter the equation I've written above \[y = -\frac{(x -3)^2}{4} + 7\]

  15. anonymous
    • one year ago
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    y = -0.25x^2 + 1.5x + 4.75?

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