- anonymous

Find the standard form of the equation of the parabola with a focus at (0, -2) and a directrix at y = 2.

- schrodinger

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- anonymous

- Hero

@Nairz, what are your thoughts regarding this problem? Do you have an approach for solving this?

- 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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## More answers

- anonymous

- 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

And that will give me the equation to the parabola?

- Hero

Yes. Notice that in this case:
Focus: \((x_1, y_1) = (0,-2)\)
Directrix: \((x_2, y_2) = (x,2)\)

- Hero

You'll have to do a bit of simplification after inserting the points in order to express the equation in standard form.

- anonymous

yes. Thanks!

- Hero

@Nairz, mind showing your work for this? Let's see what equation you come up with.

- anonymous

ummm I got \[x=2\sqrt{2}\]

- anonymous

ill show my work in a second

- anonymous

\[\large (x-0)^2 + (y+2)^2 = (x-x)^2 + (y-2)^2\]

- anonymous

\[\large x^2 + (y+2)^2 = (y+2)^2\]

- anonymous

\[\large x^2 = (y-2)^2 - (y+2)^2\]

- anonymous

\[\large x^2 = -8\]

- anonymous

\[\large x=2\sqrt{2}\]

- Hero

Okay, hang on a second.

- 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

i dont have a formula sheet in front of me

- Hero

Difference of squares formula:
\(a^2 - b^2 = (a + b)(a - b)\)

- Hero

In this case
a = y - 2
b = y + 2

- anonymous

so its \[\large ((y-2)+ (y+2))((y-2)-(y+2))\]

- Hero

Don't forget the \(x^2\) equals part

- anonymous

simplified: \[\large x^2 = 2y(-4)\]

- anonymous

so x^2 = -8y

- anonymous

which isnt an option...

- Hero

Finish isolating y

- anonymous

y=x^2/-8

- Hero

Note that the equation can also be written in the form
\(y = -\dfrac{1}{8}x^2\)

- anonymous

Ok... I just realized I was looking at the wrong problem's answers... That is there. Thanks

- anonymous

Can you work through another one of these with me? Just to make sure I do it correctly?

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