anonymous
  • anonymous
HELP PLEASE ON PRECALC, Here's the question, help quickly and correctly and you shall receive a medal! lol: Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, -7).
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
@phi
anonymous
  • anonymous
Hello?
anonymous
  • anonymous
Please help, I need to know how to do this, I am working alone online - my teacher won't help and the lesson doesn't make sense. Please help.

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anonymous
  • anonymous
@phi can you help?
phi
  • phi
First, plot the focus and vertex (roughly , need not be a work of art) can you do that?
anonymous
  • anonymous
sure hold on
anonymous
  • anonymous
actually no, can you? not sure where everything goes
phi
  • phi
vertex is (0,0) and focus is (0,-7) can you plot these points?
anonymous
  • anonymous
yep|dw:1437595113760:dw|
phi
  • phi
and the parabola is a U shape the "curls around" the focus the vertex is the "tip" of the parabola. so it looks like this |dw:1437595121260:dw|
anonymous
  • anonymous
ok so how do we get the equation now?
anonymous
  • anonymous
uhoh
phi
  • phi
the standard equation for the simplest parabola (with vertex at 0,0) is y= x^2 this one is upside down, so expect it to be y = - a x^2 to find y, we use the idea that 4 p y = x^2 is the equation of a parabola and "p" is the distance from the vertex to the focus
phi
  • phi
in this case, the distance from -7 to 0 is 7 (ignoring the sign) 4p is 4*7= 28 so 28y= x^2 y = 1/28 x^2 but it has to be negative because we are upside down so \[ y = - \frac{1}{28} x^2\]
anonymous
  • anonymous
so for p i do this: \[\left| -7 - 0 \right| = 7 so p = 7\]
phi
  • phi
that works
anonymous
  • anonymous
oh ok. thanks can you help with one more. same type of problem, different info given
anonymous
  • anonymous
Here it is: Find the standard form of the equation of the parabola with a focus at (7, 0) and a directrix at x = -7.
anonymous
  • anonymous
i am going to draw it is that ok?
phi
  • phi
yes, just to get an idea of what is going on.
anonymous
  • anonymous
|dw:1437595918471:dw| it looks bad but i think that is the right idea?
anonymous
  • anonymous
so what equation do i use for horizontal parabolas
phi
  • phi
ok, except the vertex is exactly halfway between the directrix and focus
anonymous
  • anonymous
oh ok so...
phi
  • phi
you can just average the x values (-7+7)/2 = 0 the vertex is at (0,0) this one is sideways, so swap x and y in the standard 4py= x^2 to get 4p x = y^2 just like before p is the distance from the vertex to the focus
anonymous
  • anonymous
which is 7 right? so p =7
anonymous
  • anonymous
4(7)x = y^2?
anonymous
  • anonymous
which simplifies to 28x = y^2, but that's not an answer choice. do you want to see them?
phi
  • phi
ok
anonymous
  • anonymous
y = (1/28)x^2 x = (1/28)y^2 -28y = x^2 y^2 = 14x
phi
  • phi
28x = y^2 divide both sides by 28
anonymous
  • anonymous
oh ok awesome! thanks so much!
phi
  • phi
yw

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