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).

- anonymous

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- anonymous

@phi

- anonymous

Hello?

- 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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## More answers

- anonymous

@phi can you help?

- phi

First, plot the focus and vertex
(roughly , need not be a work of art)
can you do that?

- anonymous

sure hold on

- anonymous

actually no, can you? not sure where everything goes

- phi

vertex is (0,0) and focus is (0,-7)
can you plot these points?

- anonymous

yep|dw:1437595113760:dw|

- 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

ok so how do we get the equation now?

- anonymous

uhoh

- 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

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

so for p i do this:
\[\left| -7 - 0 \right| = 7 so p = 7\]

- phi

that works

- anonymous

oh ok. thanks can you help with one more. same type of problem, different info given

- 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

i am going to draw it is that ok?

- phi

yes, just to get an idea of what is going on.

- anonymous

|dw:1437595918471:dw|
it looks bad but i think that is the right idea?

- anonymous

so what equation do i use for horizontal parabolas

- phi

ok, except the vertex is exactly halfway between the directrix and focus

- anonymous

oh ok so...

- 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

which is 7 right? so p =7

- anonymous

4(7)x = y^2?

- anonymous

which simplifies to 28x = y^2, but that's not an answer choice. do you want to see them?

- phi

ok

- anonymous

y = (1/28)x^2
x = (1/28)y^2
-28y = x^2
y^2 = 14x

- phi

28x = y^2
divide both sides by 28

- anonymous

oh ok awesome! thanks so much!

- phi

yw

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