Question

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.

Answer 1

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

Answer 2

@jchick

Answer 3

Sorry it took so long to reply lol

Answer 4

its cool XD

Answer 5

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

Answer 6

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.

Answer 7

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

Answer 8

but how do you find (x,y)?

Answer 9

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

Answer 10

why?

Answer 11

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

Answer 12

oh okay, that makes sense.

Answer 13

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

Answer 14

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

Answer 15

Does this make sense so far?

Answer 16

im not sure... itdoes a little bit

Answer 17

Ok where do you need explanation?

Answer 18

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

Answer 19

sorry that is the correct version

Answer 20

what are you plugging them into?

Answer 21

What do you mean?

Answer 22

Where?

Answer 23

like are you plugging those into an equation??

Answer 24

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

Answer 25

Yes

Answer 26

okay, what equation?

Answer 27

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

Answer 28

im so confused....

Answer 29

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

Answer 30

Do you understand now?

Answer 31

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

Answer 32

Oh ok

Answer 33

Sorry for confusion

Answer 34

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

Answer 35

Having internet trouble please hang on a second

Answer 36

okay

Answer 37

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

Answer 38

so do you use the distance formula to find a?

Answer 39

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

Answer 40

Sorry I am back and thank you @peachpi

Answer 41

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

Answer 42

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

Answer 43

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