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.

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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.

Mathematics
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I don't quite understand how to find a when writing the equation.. that's all I need help with.
Sorry it took so long to reply lol

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Other answers:

its cool XD
I waited till I saw you weren't actively helping someone
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.
(x+2)^2+(y−5)^2=(y−13)^2 Simplify above.
but how do you find (x,y)?
no matter what value x is, y will always be 13
why?
directrix y = 13 can be represented as the point (x, 13)
oh okay, that makes sense.
So after inputting the points, you have: (x−(−2))2+(y−5)2−−−−−−−−−−−−−−−−−−√=(x−x)2+(y−13)2−−−−−−−−−−−−−−−−√
You can then square both sides to get: (x−(−2))2+(y−5)2= (x−x)2+(y−13)2
Does this make sense so far?
im not sure... itdoes a little bit
Ok where do you need explanation?
You can then square both sides to get: (x−(−2))^2+(y−5)^2= (x−x)^2+(y−13)^2
sorry that is the correct version
what are you plugging them into?
What do you mean?
Where?
like are you plugging those into an equation??
And then that simplifies to (x+2)^2+(y−5)^2=(y−13)^2
Yes
okay, what equation?
(x−(−2))2+(y−5)2−−−−−−−−−−−−−−−−−−√= (x−x)2+(y−13)2−−−−−−−−−−−−−−−−√
im so confused....
inserted points (-2,5) and (x, 13) the distance formula: (x−x1)2+(y−y2)2−−−−−−−−−−−−−−−−−√=(x−x2)2+(y−y2)2−−−−−−−−−−−−−−−−−√
Do you understand now?
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
Oh ok
Sorry for confusion
its cool I just didn't know what you were plugging it all into.. XD
Having internet trouble please hang on a second
okay
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
so do you use the distance formula to find a?
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
Sorry I am back and thank you @peachpi
just FOIL out the equation and you'll have an expanded equation for the parabola
im still really confused... what is r in that equation??
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|
how do you find the vertex?
(x-a)^2 =-4p(y-b) vertex =(a,b)
@kaylamarie79 do you remember the standard form?
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.
y=ax^2+bx+c
@peachpi I dont know p
p is the focal distance
p = distance from vertex to focus = distance from vertex to directrix = half the distance from focus to directrix
1 min im trying wrap my mind around this..
what are you thinking?
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...
ok. starting from the beginning. The directrix is y = 13. The focus is (-2, 5).|dw:1439945735884:dw|
uh- huh
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|
how do you get the vertex point?
what number is halfway between 5 and 13?
oh okay. how do you know the -2 though?
because the parabola opens down, the focus and vertex must have the same x-coordinate
is it always equal as the focus or only when opening down?
only when opening up or down. When it opens left or right the y-coordinates are the same
oh okay that makes sense
that's the vertex. Now for p. That's the distance between the focus and vertex, so it's p = 9 - 5.
p = 4
with me?
yeah, so to get a you use the focal length equation but just plug in p?
yes. but since it opens down make p be -4 and you can drop the absolute value signs
ohhhhh see the other ways just got really confusing..
okk cool
Did you find your answer?
yeah, I did.., I understand now XD
Ok good
thank you guys for helping
Yeah no problem and sorry about being confusing lol

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