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Conics |
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| Nov5-08, 08:41 PM | #1 |
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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 |
| 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.
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| Nov5-08, 10:41 PM | #3 |
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when i completed the square i got [tex](x-1)^2 = 8(-y+\frac{7}{8})[/tex]
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| 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?
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| Nov5-08, 11:02 PM | #5 |
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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?
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| Nov5-08, 11:10 PM | #7 |
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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.
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| Nov5-08, 11:19 PM | #9 |
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Ok thanks for all of your help!
I got: v = (1, 1) p = 2 |
| Nov6-08, 09:23 PM | #10 |
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So does that look right? If so, where do I go from here?
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| Nov6-08, 09:34 PM | #11 |
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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]
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| Nov6-08, 09:46 PM | #12 |
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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?
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| Nov6-08, 09:58 PM | #14 |
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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.
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| Nov6-08, 10:15 PM | #16 |
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the parabola opens up.
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| Nov6-08, 10:18 PM | #17 |
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