anonymous
  • anonymous
Find the standard form of the equation of the parabola with a focus at (0, -10) and a directrix at y = 10
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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jdoe0001
  • jdoe0001
keeping in mind that the distance from the vertex to the focus, is the exact same one as the distance from the vertex to the directrix, in the opposite direction |dw:1378247017716:dw| where do you think is the vertex?
anonymous
  • anonymous
is it 10?
jdoe0001
  • jdoe0001
well, if the distance from the vertex to the focus, is the exact same one as the distance from the vertex to the directrix that means that the vertex is half-way between both of them

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anonymous
  • anonymous
So 5
jdoe0001
  • jdoe0001
|dw:1378247574330:dw|
jdoe0001
  • jdoe0001
see now where the vertex is?
anonymous
  • anonymous
0?
jdoe0001
  • jdoe0001
yeap, is at the origin, 0, 0 notice the distance from the vertex to the focus (0, -10), is 10 units the "focus form" for a parabola like so is \(\bf (x-h)^2=4p(y-k)\) (h, k) = vertex coordinates p = distance from the vertex to the focus point
anonymous
  • anonymous
so i just plug the numbers in? (0 - 0)^2 = 4(10) (-10 - 0)
jdoe0001
  • jdoe0001
well, \(\bf (x-h)^2=4p(y-k) \implies (x-0)^2=4(10)(y-0)\)
jdoe0001
  • jdoe0001
yes, and then you can solve for "y" or equate to 0
jdoe0001
  • jdoe0001
well, it has an "x" and "y", so might as well solve for "y"
anonymous
  • anonymous
x^2 - 40 = y
jdoe0001
  • jdoe0001
well, \(\bf \bf (x-h)^2=4p(y-k) \implies (x-0)^2=4(10)(y-0) \implies x^2 = 40y\\ \cfrac{x^2}{40} = y \implies \cfrac{1}{40}x^2 = y \)
jdoe0001
  • jdoe0001
well, that's the parabola equation
anonymous
  • anonymous
Oh okay. thank you!
jdoe0001
  • jdoe0001
yw

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