Question

Write the equations in standard form (I'm working with parabolas):
x=-1/8 y^2
x-5=4(y-3)^2
x^2-2y-6x=-5
y=-x^2

Answer 1

do you know what the standard form for a parabola is?

Answer 2

In my notes I have different equations for horizontal and vertical parabolas but I don't know what the standard form for the parabola is. @blurbendy

Answer 3

standard form for a parabola is y = ax^2 + bx + c

Answer 4

Okay. I'm sure I can figure out how to put the equations into that form, but how do I figure out the focal point and directrix from that equation?

Answer 5

@blurbendy

Answer 6

it's been a while since I've done these, so I don't quite remember.

I should also mention that if a parabola is "sideways" then the equation is x = ay^2 + by + c

Answer 7

Okay, thank you

Answer 8

sorry, guess I need to retake algebra 2 XD

Answer 9

actually, I just looked it up. it's not that hard. let's start with the last one.

y = -x^2

form: x^2 = (4p)y

y = -p (directrix)

focus: (0, p)

step 1: get -x^2 by itself

-y = x^2

step 2:

4p = -1

p = -1/4

y = -p, y = 1/4, directrix = 1/4

focus = (0, -1/4)

Answer 10

let me know if you have any questions

Answer 11

@blurbendy For the 3rd equation how would I get it into a form so I can figure out what p is?

Answer 12

youll want to get it in y = ax^2 + bx + c form

Answer 13

get y by itself

Answer 14

so it would be y = 1/2(x^2 - 6x +5) ? and then to find p it would be 4p = 1/2 ?

Answer 15

the standard form looks correct, but Im not sure how to find p