anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I don't quite understand how to find a when writing the equation.. that's all I need help with.
anonymous
  • anonymous
@jchick
jchick
  • jchick
Sorry it took so long to reply lol

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

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