anonymous
  • anonymous
Find the standard form of the equation of the parabola with a focus at (0, -2) and a directrix at y = 2.
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
Hero
  • Hero
@Nairz, what are your thoughts regarding this problem? Do you have an approach for solving this?
anonymous
  • anonymous
Im assuming the vertex is the origin since the directrix and the focus are both 2 units away from the origin along the same line.

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anonymous
  • anonymous
Hero
  • Hero
Consider this: If you have two points, the focus \((x_1, y_1)\) and the directrix \((x_2, y_2)\) then you can insert those points in to the following formula to find the standard form of the equation of a parabola: \((x - x_1)^2 + (y - y_1)^2 = (x - x_2)^2 + (y - y_2)^2\)
anonymous
  • anonymous
And that will give me the equation to the parabola?
Hero
  • Hero
Yes. Notice that in this case: Focus: \((x_1, y_1) = (0,-2)\) Directrix: \((x_2, y_2) = (x,2)\)
Hero
  • Hero
You'll have to do a bit of simplification after inserting the points in order to express the equation in standard form.
anonymous
  • anonymous
yes. Thanks!
Hero
  • Hero
@Nairz, mind showing your work for this? Let's see what equation you come up with.
anonymous
  • anonymous
ummm I got \[x=2\sqrt{2}\]
anonymous
  • anonymous
ill show my work in a second
anonymous
  • anonymous
\[\large (x-0)^2 + (y+2)^2 = (x-x)^2 + (y-2)^2\]
anonymous
  • anonymous
\[\large x^2 + (y+2)^2 = (y+2)^2\]
anonymous
  • anonymous
\[\large x^2 = (y-2)^2 - (y+2)^2\]
anonymous
  • anonymous
\[\large x^2 = -8\]
anonymous
  • anonymous
\[\large x=2\sqrt{2}\]
Hero
  • Hero
Okay, hang on a second.
Hero
  • Hero
\((y - 2)^2 - (y + 2)^2\) is actually "difference of squares". You should use the difference of squares formula to simplify that properly.
anonymous
  • anonymous
i dont have a formula sheet in front of me
Hero
  • Hero
Difference of squares formula: \(a^2 - b^2 = (a + b)(a - b)\)
Hero
  • Hero
In this case a = y - 2 b = y + 2
anonymous
  • anonymous
so its \[\large ((y-2)+ (y+2))((y-2)-(y+2))\]
Hero
  • Hero
Don't forget the \(x^2\) equals part
anonymous
  • anonymous
simplified: \[\large x^2 = 2y(-4)\]
anonymous
  • anonymous
so x^2 = -8y
anonymous
  • anonymous
which isnt an option...
Hero
  • Hero
Finish isolating y
anonymous
  • anonymous
y=x^2/-8
Hero
  • Hero
Note that the equation can also be written in the form \(y = -\dfrac{1}{8}x^2\)
anonymous
  • anonymous
Ok... I just realized I was looking at the wrong problem's answers... That is there. Thanks
anonymous
  • anonymous
Can you work through another one of these with me? Just to make sure I do it correctly?

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