anonymous
  • anonymous
Find the standard form of the equation of the parabola with a focus at (0, 6) and a directrix at y = -6. PLEASE HELP ME
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
take any point (x,y) and then use the definition of parabola.
anonymous
  • anonymous
what @surjithayer
anonymous
  • anonymous
Let P (x,y) be any point on the parabola. by definition PM=PF

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jdoe0001
  • jdoe0001
keep in mind that the directrix in a parabola, is equally distant from the vertex as the focus, if the focus is at (0, 6) and the directrix is at y=-6 [ a flat horizontal line down below ], the vertex is half-way between them also recall that a parabola "focus form" equation for a vertical opening parabola is \(\bf (x-h)^2=4p(y-k)\) (h, k) = coordinates of the vertex p = distance from the vertex to the focus/directrix
anonymous
  • anonymous
|dw:1378075415692:dw|
anonymous
  • anonymous
so I would have (x-h)^2=4*6(-6-k)?
jdoe0001
  • jdoe0001
close, yes, notice @surjithayer 's graph, where the origin is at, so that'd be your (h, k) values
anonymous
  • anonymous
wait so how I know what h and k is?? I'm still a bit confused on how to read that to get h,k
anonymous
  • anonymous
@surjithayer @jdoe0001
anonymous
  • anonymous
by definition PM=PF \[PM=\frac{ \left| y+6 \right| }{\sqrt{0^{2}+1^{2}} }=\left| y+6 \right|\] \[PF=\sqrt{\left( x-0 \right)^{2}+\left( y-6 \right)^{2}}\] \[\left| y+6 \right|=\sqrt{x ^{2}+\left( y-6 \right)^{2}}\] squaring both sides and solve you get the required equation.
jdoe0001
  • jdoe0001
well, based on the half-way distance of (0, 6) and y= -6, as shown in the graph by surjithayer, the vertex is the origin
anonymous
  • anonymous
so it'd be 3?
jdoe0001
  • jdoe0001
3? for ... ?
anonymous
  • anonymous
for h,k? it be 3,3?
jdoe0001
  • jdoe0001
|dw:1378076989399:dw|
jdoe0001
  • jdoe0001
the parabola opens up TOWARDS the focus, and its vertex is half-way between the focus and the directrix
anonymous
  • anonymous
\[y ^{2}+12y+36=x ^{2}+y ^{2}-12y+36, or x ^{2}=24y\]
jdoe0001
  • jdoe0001
(h, k) = (0, 0) p = 6 \(\bf (x-h)^2=4p(y-k) \implies (x-0)^2=4(6)(y-0)\)
anonymous
  • anonymous
so in other words the answer would be y=(1/24)x^2 once simplified?
jdoe0001
  • jdoe0001
yeap

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