Regan is trying to find the equation of a quadratic that has a focus of (−2, 5) and a directrix of y = 13. Describe to Regan your preferred method for deriving the equation. Make sure you use Regan's situation as a model to help her understand.

- anonymous

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- anonymous

I don't quite understand how to find a when writing the equation.. that's all I need help with.

- anonymous

@jchick

- jchick

Sorry it took so long to reply lol

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## More answers

- anonymous

its cool XD

- anonymous

I waited till I saw you weren't actively helping someone

- jchick

The definition of a parabola is that the distance of any point on it from the focus is equal to the distance of the point from the directrix. So if (x,y) is a point on the parabola, then find the distance of (x,y) from the focus (-2,5). Find the distance of (x,y) from y = 13. Equate them and you will have your quadratic equation.

- jchick

(x+2)^2+(y−5)^2=(y−13)^2
Simplify above.

- anonymous

but how do you find (x,y)?

- jchick

no matter what value x is, y will always be 13

- anonymous

why?

- jchick

directrix y = 13 can be represented as the point (x, 13)

- anonymous

oh okay, that makes sense.

- jchick

So after inputting the points, you have: (x−(−2))2+(y−5)2−−−−−−−−−−−−−−−−−−√=(x−x)2+(y−13)2−−−−−−−−−−−−−−−−√

- jchick

You can then square both sides to get: (x−(−2))2+(y−5)2= (x−x)2+(y−13)2

- jchick

Does this make sense so far?

- anonymous

im not sure... itdoes a little bit

- jchick

Ok where do you need explanation?

- jchick

You can then square both sides to get: (x−(−2))^2+(y−5)^2= (x−x)^2+(y−13)^2

- jchick

sorry that is the correct version

- anonymous

what are you plugging them into?

- jchick

What do you mean?

- jchick

Where?

- anonymous

like are you plugging those into an equation??

- jchick

And then that simplifies to (x+2)^2+(y−5)^2=(y−13)^2

- jchick

Yes

- anonymous

okay, what equation?

- jchick

(x−(−2))2+(y−5)2−−−−−−−−−−−−−−−−−−√=
(x−x)2+(y−13)2−−−−−−−−−−−−−−−−√

- anonymous

im so confused....

- jchick

inserted points (-2,5) and (x, 13) the distance formula: (x−x1)2+(y−y2)2−−−−−−−−−−−−−−−−−√=(x−x2)2+(y−y2)2−−−−−−−−−−−−−−−−−√

- jchick

Do you understand now?

- anonymous

like I was learning it like this... The graph of the quadratic is a parabola.
The directrix is horizontal, so the parabola is vertical.
The focus lies below the directrix, so the parabola opens downwards.
General equation of a down-opening parabola:
y = a(x - h)² + k
with
a < 0
vertex (h, k)
focal length p = | 1/(4a) |
focus (h, k-p)
directrix y = k+p
Plug your data into the equations and solve for a
now im really confused... when I was only confused about how to figure out what a is XD

- jchick

Oh ok

- jchick

Sorry for confusion

- anonymous

its cool I just didn't know what you were plugging it all into.. XD

- jchick

Having internet trouble please hang on a second

- anonymous

okay

- anonymous

I've never seen this done this way. Pretty cool @jchick
It looks like this is using the definition of a parabola as the set of points equidistant between the focus and directrix. A circle is a set of point equidistant from the same point.
The equation for a circle is
\[(x-h)^2+(y-k)^2=r^2\].
jchick set the equations for 2 circles equal to each other, one for the focus with (-2, 5) and the other for the directrix (x, 13). That's where this came from
\[(x+2)^2+(y-5)^2=(x-x)^2+(y-13)\]
If you mulitply everything out, and then complete the square you can find a with this method

- anonymous

so do you use the distance formula to find a?

- anonymous

that's pretty much what it works out to being. and you don't actually need to complete the square. a will be the coefficient of x²

- jchick

Sorry I am back and thank you @peachpi

- anonymous

just FOIL out the equation and you'll have an expanded equation for the parabola

- anonymous

im still really confused... what is r in that equation??

- anonymous

Maybe try a different method since this is confusing. We know that the vertex is halfway between the focus and directrix, so it's (-2, 9) and the parabola opens down like you said above. So the focal length is 9 - 5 = 4
|dw:1439944412409:dw|

- anonymous

how do you find the vertex?

- jchick

(x-a)^2 =-4p(y-b) vertex =(a,b)

- jchick

@kaylamarie79 do you remember the standard form?

- anonymous

The parabola opens down so the vertex has to be directly above the focus, meaning the x-coordinate has to be -2. The vertex is halfway between focus and directrix, so it's halfway between 5 and 13, so the y-coordinate has to be 9. Vertex = (-2, 9).
To find a:
From your list of equations above,
\[p=\left| \frac{ 1 }{ 4a } \right|\]
Since you know p, you can solve for a.

- anonymous

y=ax^2+bx+c

- anonymous

@peachpi I dont know p

- anonymous

p is the focal distance

- anonymous

p = distance from vertex to focus = distance from vertex to directrix = half the distance from focus to directrix

- anonymous

1 min im trying wrap my mind around this..

- anonymous

what are you thinking?

- anonymous

all I can think is i'm sooo confused... like I don't understand where your getting a the focal point(p) or the vertex... and I keep getting more confused as y'all repeat things.. like i'm terrible in math and have to see every single step or the entire equation gets jumbled around in my head...

- anonymous

ok. starting from the beginning. The directrix is y = 13. The focus is (-2, 5).|dw:1439945735884:dw|

- anonymous

uh- huh

- anonymous

The vertex has to be halfway between them. 9 is halfway between 5 and 13, so the vertex has to be (-2, 9)|dw:1439945891758:dw|

- anonymous

how do you get the vertex point?

- anonymous

what number is halfway between 5 and 13?

- anonymous

oh okay. how do you know the -2 though?

- anonymous

because the parabola opens down, the focus and vertex must have the same x-coordinate

- anonymous

is it always equal as the focus or only when opening down?

- anonymous

only when opening up or down.
When it opens left or right the y-coordinates are the same

- anonymous

oh okay that makes sense

- anonymous

that's the vertex. Now for p. That's the distance between the focus and vertex, so it's
p = 9 - 5.

- anonymous

p = 4

- anonymous

with me?

- anonymous

yeah, so to get a you use the focal length equation but just plug in p?

- anonymous

yes. but since it opens down make p be -4 and you can drop the absolute value signs

- anonymous

ohhhhh see the other ways just got really confusing..

- anonymous

okk cool

- jchick

Did you find your answer?

- anonymous

yeah, I did.., I understand now XD

- jchick

Ok good

- anonymous

thank you guys for helping

- jchick

Yeah no problem and sorry about being confusing lol

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