Algebra- conics- y^2 + 4y - x^2 + 2x = 6

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burnt_umber
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$1.00 Algebra- conics- y^2 + 4y - x^2 + 2x = 6

From Mathematics: Algebra
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Due on Nov. 17, 2008
Asked on Nov 15, 2008 at 1:22:02PM

Q:
Sketch the graph and give the eccentricity. If a parabola, lable focus and directrix. If a circle, give center and radius. If ellipse give lengths of the major and minor axes and location of foci. If hyperbola, give the vertices, foci, and slopes of asymptotes.

y^2 + 4y - x^2 + 2x = 6

Please explain clearly- like you're talking to a 6 year old. I'm having a hard time understanding this stuff.

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b_h

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$1.00 Conic Sketching and Analysis

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Posted on Nov 15, 2008 at 3:35:39PM

A:
Preview: ... the x^2 term is negative we can put it in the form

(y - k)^2 / a^2 - (x - h)^2 / b^2 = 1

This is a north-south opening hyperbola, or a hyperbola with a vertical transverse axis.

We'll do this in a series of steps.

First we'll rewrite the equation in the form
(y^2 + 4y) - (x^2 - 2x) = 6

This is just group the terms.

Now, we want to complete the square of y^2 + 4y. If we take the 4, divide by 2 and then square we get 4. So adding 4 we get: y^2 + 4y + 4 = (y + 2)^2 or y^2 + 4y = (y + 2)^2 - 4.

Plugging this into our equation we have

(y ...

The full tutorial is about 618 words long plus attachments.

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Sketch.pdf (37K) (Preview)

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Xcel
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$1.00 Answer and explanation step by step

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Posted on Nov 15, 2008 at 3:52:17PM

A:
Preview: ... br>This means that -a or a distance away from the center are the vertices.

The center is (1,-4) (you get from graph/equation)
Therefore the vertices are (1,-4+sqrt(11)) and (1,-4+sqrt(11))

The foci are found using the equation (h+ or minus c,k)
h is the x coordinate for center and k is the y coordinate

Also
c = sprt(a^2+b^2)

w ...

The full tutorial is about 239 words long plus attachments and additional clarificationclarifications added later.

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Graph.gif (1K)