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- Aug 3rd 2009, 09:48 PMLAZER117Need urgent help with some problems!!
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- Aug 3rd 2009, 09:55 PMyeongil
That's more than a

**couple**of problems. Having that many problems in one thread is generally frowned upon here.

The vertex form of the parabola is

$\displaystyle (y - k)^2 = 4p(x - h)$,

where (h, k) is the vertex and p is the focal length. Since the y-term is squared, the parabola opens either to the left or to the right, which means that the focus would be at point

$\displaystyle (h + p, k)$

and the directrix would be the equation

$\displaystyle x = h - p$.

Quote:

Find the center, foci, and vertices of the ellipse.

4x2 + 5y2 - 24x + 60y + 196 = 0

$\displaystyle \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$ OR

$\displaystyle \frac{(y - k)^2}{a^2} + \frac{(x - h)^2}{b^2} = 1$.

See this site for more info: Conics: Ellipses: Introduction .

To make the problem look like one of the two forms above, you'll need to complete the square.

$\displaystyle 4x^2 + 5y^2 - 24x + 60y + 196 = 0$

Rearrange the terms first.

$\displaystyle 4x^2 - 24x + 5y^2 + 60y = -196$

Factor out the common factor:

$\displaystyle 4(x^2 - 6x) + 5(y^2 + 12y) = -196$

From here, complete the square. (I assume you know how to do this.) Then divide both sides by the number you'll end up with on the right side. The center is at (h, k). Consult the site I linked above to find the vertices and foci.

Your hyperbola equation is missing.

01 - Aug 3rd 2009, 10:07 PMLAZER117
Sorry for all the problems posted , but I haven't been able to go to class due to sickness.

I really appreciate your help, and things seem much clearer now.

Hopefully I'll stop being a newbie :P

Thanks again. - Aug 3rd 2009, 10:15 PMearboth
The stationary vector of P is $\displaystyle \vec p = \langle-6,-3\rangle$ and the staionary vector of Q is $\displaystyle \vec q = \langle -1, 1\rangle$.

Then the vector $\displaystyle \overrightarrow{PQ} = \vec q - \vec p$

Quote:

If P = (6, 10) and Q = (x, 106), find all numbers x such that the vector represented by has length 120.

2. Calculate the length of the vector: If $\displaystyle \vec v = \langle a, b \rangle$ then $\displaystyle |\vec v| = \sqrt{a^2+b^2}$

3. In your case you'll get: $\displaystyle |\overrightarrow{PQ}|=\sqrt{(x-6)^2+(-96)^2} = 120$

4. Solve for x.

Quote:

Find angle between v and w.

**v**= -3**i**+ 4**j,****w**= -2**i**+ 9**j**

Quote:

Which of the following vectors is parallel to**v =**12**i**+ 30**j**?

http://myutbtsc.blackboard.com/images/spacer.gif

**w**= - 20**i**- 50**j**

**w**= 2**i**+ 6**j**

**w**= 16**i**+ 20[B]j

Since $\displaystyle \vec v = 6 \cdot \langle 2i+5j \rangle$ you should look for a vector $\displaystyle \vec w$ whose components are in the ratio 2 : 5.

Quote:

Find the vertex, focus, and directrix of the parabola with the given equation.

$\displaystyle (y + 3)^2 = 8(x - 1)$

Quote:

Find the center, foci, and vertices of the ellipse.

$\displaystyle 4x^2 + 5y^2 - 24x + 60y + 196 = 0$

Quote:

Find the asymptotes of the hyperbola.

- = 1

- Aug 3rd 2009, 11:45 PMLAZER117
I've read the links you have provided and I'm not having trouble anymore with those. I'm still stumped on:

If P = (6, 10) and Q = (x, 106), find all numbers x such that the vector represented by PQ has length 120. - Aug 4th 2009, 12:00 AMsongoku
Hi LAZER117

Hint: let x (a, b) and Y (c,d), then the distance between X and Y = $\displaystyle \sqrt{(c-a)^2+(d-b)^2}$