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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
$\displaystyle x = h - p$.
The equation for the ellipse is usually written in one of 2 forms: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
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 $\displaystyle \overrightarrow{PQ} = \vec q - \vec p$
1. Find the vector $\displaystyle \overrightarrow{PQ}$
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.
Use the formula: $\displaystyle \cos(\angle(\vec v, \vec w))=\dfrac{\vec v \cdot \vec w}{|\vec v| \cdot |\vec w|}$Find angle between v and w.
v = -3i + 4j,w = -2i + 9j
You have to find one single factor which transforms the vector $\displaystyle \vec v $ into one of the vectors $\displaystyle \vec w$.Which of the following vectors is parallel to v = 12i + 30j?
w = - 20i - 50j
w = 2i + 6j
w = 16i + 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.
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 vertex, focus, and directrix of the parabola with the given equation.
$\displaystyle (y + 3)^2 = 8(x - 1)$
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.
$\displaystyle 4x^2 + 5y^2 - 24x + 60y + 196 = 0$
Seems to be the condensed form of a hyperbola
Find the asymptotes of the hyperbola.
- = 1