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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

its cool XD

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

(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.

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....

Do you understand now?

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

so do you use the distance formula to find a?

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??

how do you find the vertex?

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

@kaylamarie79 do you remember the standard form?

y=ax^2+bx+c

p is the focal distance

1 min im trying wrap my mind around this..

what are you thinking?

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

uh- huh

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