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Conics

 
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Nov5-08, 08:41 PM   #1
 

Conics


1. The problem statement, all variables and given/known data

Graph the following. Include center, verticies, foci, asymptote, and directrix as appropriate.

2. Relevant equations

[tex]x^2 + 8y - 2x = 7[/tex]

3. The attempt at a solution

So far I have:

V = (1, -7/8)
P = -2
X = -1

I have no clue where to go from here, or if I'm even right.
Thanks
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Nov5-08, 09:50 PM   #2

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Don't you have to decide what kind of conic it is first? Complete the square in x and try to write it in some kind of normal form.
Nov5-08, 10:41 PM   #3
 
when i completed the square i got [tex](x-1)^2 = 8(-y+\frac{7}{8})[/tex]
Nov5-08, 10:48 PM   #4

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Conics


I think you mean (x-1)^2=8*(1-y). Try that once more. It's a parabola, isn't it?
Nov5-08, 11:02 PM   #5
 
why would it be 8(y-1) ? What happened to the 7?
How do you tell that it's a parabola? Because of the (x-1)^2?
Nov5-08, 11:04 PM   #6

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Yes, because it's quadratic in x and linear in y. The 7 changed to an 8 when you added the 1 to both sides to complete the square. You did do that, right?
Nov5-08, 11:10 PM   #7
 
oh nooes. I didn't.

ok i got it. Now what?
Nov5-08, 11:14 PM   #8

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Ok, now where is the vertex? And if you tell me what X and P are supposed to be I might be able to help you with those. Tomorrow. zzzzzzzzzzz.
Nov5-08, 11:19 PM   #9
 
Ok thanks for all of your help!

I got:

v = (1, 1)
p = 2
Nov6-08, 09:23 PM   #10
 
So does that look right? If so, where do I go from here?
Nov6-08, 09:34 PM   #11
 
really I think my problem is putting the initial equation in the [tex]\frac{(y-y0)^2}{a^2}-\frac{(x-x0)^2}{b^2}[/tex]
Nov6-08, 09:46 PM   #12
 
Mistake. I know it's a parabola because it's in the form [tex](x-1)^2=8(1-y)[/tex]. And I know p is 2 and the vertex is at (1,1). But how do I know which direction from V to go two units? Up or down? Left or right?

Also how do I find the asymptote?
Nov6-08, 09:51 PM   #13

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You never told me what P is. Is it the distance from the vertex to the focus? Does a parabola have any asymptotes?
Nov6-08, 09:58 PM   #14
 
P = 2 right?
Right now I have:

[tex]v=(1,1)[/tex]
[tex]p=2[/tex]
[tex]F=(1,3)[/tex]
[tex]Directrix=(1,-2)[/tex]

Does that look right?
Nov6-08, 10:07 PM   #15

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You've got the vertex. Now does the parabola go up or down from the vertex? As x gets large does y increase to +infinity or -infinity? And the directrix is a line, not a point.
Nov6-08, 10:15 PM   #16
 
the parabola opens up.
Nov6-08, 10:18 PM   #17

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Quote by duki View Post
the parabola opens up.
I disagree. Why do you think so?
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