Question

y^2+4y-8x+4=0

(x-8)+(y+2)^2=-4

Answer 1

There was only one x... I got kinda confused. ;c

Answer 2

What are you asked to do?

Answer 3

Put it into standard form

Answer 4

@Ria23 You will need a different standard form this time. The last one we did had x^2 and y^2... since this has only one x (like you noticed) this will be a parabola standard form.

Answer 5

>~< Why are there so many, oh my goodness...

Answer 6

\(\large\color{black}{ y^2+4y-8x+4=0 }\)

\(\large\color{black}{ y^2+4y+4=8x }\)

\(\large\color{black}{ (y+2)^2=8x }\) then divide by 8 on both sides.

Answer 7

\(\large\color{black}{ x=\frac{1}{8}(y+2)^2 }\)

Answer 8

@Ria23 haha, I know. :)

Answer 9

looks like a horizontal parabola.

Answer 10

Why did the 8x get moved? .-.

Answer 11

I added 8x to both sides.

Answer 12

And then the last step, when yhu divide it, why is it 1/8 and not just an 8 in the front?

Answer 13

I divided both sides by 8.

Answer 14

@Ria23 I'm looking up some reference tables for you, are you in an Algebra II class?

Answer 15

Pre-calculus

Answer 16

OK

Answer 17

@jtvatsim in the previous question yhu helped me with, was that a hyperbola?

Answer 18

No that was a circle. But a hyperbola has a similar equation just one of the "squares" is negative and on is positive. I think I found something, are the questions you are working on from a lesson on conic sections?

Answer 19

Yes indeed. c:

Answer 20

OK, I'll upload it in a second

Answer 21

Ok. c:

Answer 22

I have to make some modifications to it so that it makes sense.

Answer 23

Alright. c: Thank yhu so much c:

Answer 24

no worries, almost done :)

Answer 25

there were a bit more of these things than I thought... :)

Answer 26

xD Yhu don't have to do this.
But it's really nice of yhu to do it, thank yhu, if yhu don't want to, yhu don't have to. Haha

Answer 27

Got it... all done! :)

Answer 28

and the original version, more detailed.

Answer 29

Thank yhu. n.n