Graph Conics: Center, Verticies, Foci, Asymptote, Directrix

[duki]

Homework Statement

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

Homework Equations

x^2 + 8y - 2x = 7

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

 

[Dick]

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.

 

[duki]

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

 

[Dick]

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

 

[duki]

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?

 

[Dick]

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?

 

[duki]

oh nooes. I didn't.

ok i got it. Now what?

 

[Dick]

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.

 

[duki]

Ok thanks for all of your help!

I got:

v = (1, 1)
p = 2

 

[duki]

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

 

[duki]

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

 

[duki]

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?

 

[Dick]

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

 

[duki]

P = 2 right?
Right now I have:

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

Does that look right?

 

[Dick]

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.

 

[duki]

the parabola opens up.

 

[Dick]

the parabola opens up.

I disagree. Why do you think so?

 

[duki]

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

 

[Dick]

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

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

 

[duki]

a - number. I got -1249

 

[Dick]

Opens up or down?

 

[duki]

opens down

 

[Dick]

Right. So where are the focus and directrix relative to the vertex? And don't give me a point for the directrix. Give me the equation for the line.

 

[duki]

the Vertex is (1,1) and the Directrix is y= -2 ?

 

[duki]

wait! Directrix is y= 3 and the Focus is (1,-1)?

 

[Dick]

Is that just a random guess? It sure doesn't show a lot of thought. If it opens down then the directrix is above the vertex and the focus is below. Come on. Look at a picture of a parabola.

 

[Dick]

wait! Directrix is y= 3 and the Focus is (1,-1)?

Much better.

 

[duki]

Ok grooovy. So are there no asymptotes? Is that only with a hyperbola?

 

[Dick]

Yes. The only conics that have asymptotes are hyperbolae.

 

[duki]

awesome. so I'm done with this one! Now I can get working on 1b) ;) Thanks for your help!I posted a thread about that one if you get a chance. I'm having a hard time getting it into standard form