# Graphing Conics help

1. Nov 5, 2008

### duki

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

$$x^2 + 8y - 2x = 7$$

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

2. Nov 5, 2008

### Dick

Re: Conics

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.

3. Nov 5, 2008

### duki

Re: Conics

when i completed the square i got $$(x-1)^2 = 8(-y+\frac{7}{8})$$

4. Nov 5, 2008

### Dick

Re: Conics

I think you mean (x-1)^2=8*(1-y). Try that once more. It's a parabola, isn't it?

5. Nov 5, 2008

### duki

Re: Conics

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?

6. Nov 5, 2008

### Dick

Re: Conics

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?

7. Nov 5, 2008

### duki

Re: Conics

oh nooes. I didn't.

ok i got it. Now what?

8. Nov 5, 2008

### Dick

Re: Conics

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.

9. Nov 5, 2008

### duki

Re: Conics

Ok thanks for all of your help!

I got:

v = (1, 1)
p = 2

10. Nov 6, 2008

### duki

Re: Conics

So does that look right? If so, where do I go from here?

11. Nov 6, 2008

### duki

Re: Conics

really I think my problem is putting the initial equation in the $$\frac{(y-y0)^2}{a^2}-\frac{(x-x0)^2}{b^2}$$

12. Nov 6, 2008

### duki

Re: Conics

Mistake. I know it's a parabola because it's in the form $$(x-1)^2=8(1-y)$$. 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?

13. Nov 6, 2008

### Dick

Re: Conics

You never told me what P is. Is it the distance from the vertex to the focus? Does a parabola have any asymptotes?

14. Nov 6, 2008

### duki

Re: Conics

P = 2 right?
Right now I have:

$$v=(1,1)$$
$$p=2$$
$$F=(1,3)$$
$$Directrix=(1,-2)$$

Does that look right?

15. Nov 6, 2008

### Dick

Re: Conics

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.

16. Nov 6, 2008

### duki

Re: Conics

the parabola opens up.

17. Nov 6, 2008

### Dick

Re: Conics

I disagree. Why do you think so?

18. Nov 6, 2008

### duki

Re: Conics

oh wait. since it's (x-1)^2 that means it goes down right?

19. Nov 6, 2008

### Dick

Re: Conics

You tell me. If x=101, what kind of number is y?

20. Nov 6, 2008

### duki

Re: Conics

a - number. I got -1249

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