Math Help - Need urgent help with some problems!!

1. Need urgent help with some problems!!

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2. That's more than a couple of problems. Having that many problems in one thread is generally frowned upon here.

Originally Posted by LAZER117
Find the vertex, focus, and directrix of the parabola with the given equation.

(y + 3)2 = 8(x - 1)
The vertex form of the parabola is
$(y - k)^2 = 4p(x - h)$,
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
$(h + p, k)$
and the directrix would be the equation
$x = h - p$.

Find the center, foci, and vertices of the ellipse.
4x2 + 5y2 - 24x + 60y + 196 = 0
The equation for the ellipse is usually written in one of 2 forms:
$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$ OR
$\frac{(y - k)^2}{a^2} + \frac{(x - h)^2}{b^2} = 1$.

To make the problem look like one of the two forms above, you'll need to complete the square.
$4x^2 + 5y^2 - 24x + 60y + 196 = 0$

Rearrange the terms first.
$4x^2 - 24x + 5y^2 + 60y = -196$

Factor out the common factor:
$4(x^2 - 6x) + 5(y^2 + 12y) = -196$

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.

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3. 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.

4. Originally Posted by LAZER117
Hey everyone, I have a couple of problems on my homework which I am having trouble solving, I would really appreciate if y'all could help me =)

The vector v has initial point P and terminal point Q. Write v in the form ai + bj; that is, find its position vector.

P = (-6, -3); Q = (-1, 1)
The stationary vector of P is $\vec p = \langle-6,-3\rangle$ and the staionary vector of Q is $\vec q = \langle -1, 1\rangle$.
Then the vector $\overrightarrow{PQ} = \vec q - \vec p$

If P = (6, 10) and Q = (x, 106), find all numbers x such that the vector represented by has length 120.
1. Find the vector $\overrightarrow{PQ}$
2. Calculate the length of the vector: If $\vec v = \langle a, b \rangle$ then $|\vec v| = \sqrt{a^2+b^2}$
3. In your case you'll get: $|\overrightarrow{PQ}|=\sqrt{(x-6)^2+(-96)^2} = 120$
4. Solve for x.

Find angle between v and w.

v = -3i + 4j,w = -2i + 9j
Use the formula: $\cos(\angle(\vec v, \vec w))=\dfrac{\vec v \cdot \vec w}{|\vec v| \cdot |\vec w|}$

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

w = - 20i - 50j
w = 2i + 6j

w = 16i + 20[B]j
You have to find one single factor which transforms the vector $\vec v$ into one of the vectors $\vec w$.
Since $\vec v = 6 \cdot \langle 2i+5j \rangle$ you should look for a vector $\vec w$ whose components are in the ratio 2 : 5.

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

$(y + 3)^2 = 8(x - 1)$
There are a lot of different methods to get these values. Could you please show us what you have done so far and where you have some difficulties, so we can help you more sufficiently.

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

$4x^2 + 5y^2 - 24x + 60y + 196 = 0$
There are a lot of different methods to get these values. Could you please show us what you have done so far and where you have some difficulties, so we can help you more sufficiently.

Find the asymptotes of the hyperbola.

- = 1
Seems to be the condensed form of a hyperbola

5. 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.

6. Hi LAZER117

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