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Math Help - vector and parabola question

  1. #1
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    vector and parabola question

    hi all I'm having a little trouble with 2 problems.

    the first is:
    with v being a vector going 3 up and 5 right, find the unit vector in the direciton of v.

    the second is:
    parabola's focus at (2,1) and vertex at (2,4). Find equation of directrix, two points where latus rectum intersects parabola, and equation of parabola itself.

    The first problem I don't know exactly where to start with what is given.

    The second problem, from what I've figured or thought to have figured is that it opens downward baesd on focus and vertex points, but not sure as to how to find the rest of the information requested.
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  2. #2
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    Quote Originally Posted by deltemis View Post
    hi all I'm having a little trouble with 2 problems.

    the first is:
    with v being a vector going 3 up and 5 right, find the unit vector in the direciton of v.

    the second is:
    parabola's focus at (2,1) and vertex at (2,4). Find equation of directrix, two points where latus rectum intersects parabola, and equation of parabola itself.

    The first problem I don't know exactly where to start with what is given.

    The second problem, from what I've figured or thought to have figured is that it opens downward baesd on focus and vertex points, but not sure as to how to find the rest of the information requested.
    Let \overrightarrow{ v^0} denote the unit vector of \vec v. Then \overrightarrow{ v^0} is defined as:

    \overrightarrow{ v^0}= \dfrac{\vec  v}{|\vec v|}

    With your values you get:

    \overrightarrow{ v^0}=\dfrac{(5, 3)}{\sqrt{34}}

    to #2:

    The axis of symmetry is a parallel of the y-axis (x-coordinates of F and V are equal) and therefore the directrix must be perpendicular to the y-axis.
    The distance between F and V is 3. Thus the directrix has the equation y = 7.

    Since F is below V the parabola opens down. The parameter p = 3 (=distance between F and V). The general equation of such a parabola is:
    -4p(y-y_V)=(x-x_V)^2

    Therefore the equation of the parabola is:

    -12(y-4)=(x-2)^2

    The endpoints of the latus rectum have the coordinates (x_F-2p\ ,\ x_F+2p). Therefore the points are at (-4, 1) and (8, 1)
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