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Math Help - Analytic Geometry

  1. #1
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    Analytic Geometry

    I am lost on these questions.

    Can somebody PLEASE help?

    1. X2=-4y Find the vertex, focus and directrix of the parabola and sketch the graph.

    2. Find the equation in standard form of the parabola with focus (-2, 4) and directrix x = 4

    3. Find the center, vertices, foci and sketch the graph of the ellipse with the following equation. 25x2+12y2=300

    Thanks for helping me keep my sanity!!

    Dave
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by Dave53 View Post
    I am lost on these questions.

    Can somebody PLEASE help?

    1. X2=-4y Find the vertex, focus and directrix of the parabola and sketch the graph.

    Dave
    x^2=-4y

    Put in vertex form: y=a(x-h)^2+k
    Vertex will be at (h, k)

    y=-\frac{1}{4}x^2

    y=-\frac{1}{4}(x-0)^2+0

    Vertex is at (0, 0)

    The equation of the directrix is  y = k-\frac{1}{4a}

    y=0-\frac{1}{4(-\frac{1}{4})}\Longrightarrow y=1

    Focus is given by (h, k+\frac{1}{4a})

    Therefore, the focus is (0, 0+\frac{1}{4(-\frac{1}{4})}\Longrightarrow (0, -1)
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  3. #3
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    Thanks for the help!

    I don't know if I'll ever understand this stuff.
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  4. #4
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    Quote Originally Posted by Dave53 View Post
    Thanks for the help!

    I don't know if I'll ever understand this stuff.
    Here:An Introduction To Conic Sections

    You can find a pretty extensive tutorial of the conic sections (definitions, examples, etc) to help you understand them better.

    I have to go now, but if you don't receive answers to your other questions you posted here, try posting them individually.

    Later.
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  5. #5
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    Quote Originally Posted by Dave53 View Post
    I am lost on these questions.

    Can somebody PLEASE help?

    1. X2=-4y Find the vertex, focus and directrix of the parabola and sketch the graph.

    2. Find the equation in standard form of the parabola with focus (-2, 4) and directrix x = 4

    3. Find the center, vertices, foci and sketch the graph of the ellipse with the following equation. 25x2+12y2=300

    Thanks for helping me keep my sanity!!

    Dave
    1. X2=-4y Find the vertex, focus and directrix of the parabola and sketch the graph.

    x^2 = -4y
    y = -(1/4)x^2 ......a basic "vertical" parabola, whose vertex is at (0,0).

    One general form of this basic vertical parabola is y = [1 /(4p)]x^2
    where p is the focal distance from the vertex.
    Also, p is the distance of the vertex from the directrix along the axis of symmetry of the parabola.

    So, 1 /(4p) = 1/4
    4p = 4
    p = 1

    Because the coefficient of the x^2 is negative...(it is -1/4 here)..., it means the parabola opens downward.

    The vertex being at (0,0), then the axis of symmetry is the x=0 vertical line, or the y-axis.

    Hence,
    >>>focus is 1 unit below the vertex, or it is at (0,-1).
    >>>directrix is 1 unit above the vertex, or it is the horizontal line y = 1.

    I don't know how to graph/draw/sketch in computers/calculators.

    --------------------------------------------
    2. Find the equation in standard form of the parabola with focus (-2, 4) and directrix x = 4

    Along the axis of symmetry, the vertex is equidistant from the focus and the directrix.

    Since the givens say that the focus is at (-2,4) and that the directrix x=4 is a vertical line, then the parabola is "horizontal"...the axis of symmetry and the directrix are perpendicular always.
    So the parabola is in the form (x -h) = [1 /(4p)](y -k)^2 ----(i)
    where
    (h,k) is the vertex
    p = focal distance from the vertex.

    We find p.
    Along the axis of symmetry, or along the horizontal line y = 4,
    absolute distance from focus to dirextrix is 2 +4 = 6 units.
    half of that is 3 units.
    That is the distance from the vertex to the focus.
    That is p. Or, p = 3 units. ------------------------**
    Also, the vertex is at point (1,4) ------------**

    Hence, the parabola is
    (x -1) = [1 /(4*3)](y -4)^2
    x -1 = (1 /12)(y -4)^2 ----------------(to be revised)

    EDIT:
    Since the directrix (x = 4) is to the right of the focus (-2,4), then the horizontal parabola opens to the left. That means the coefficient of the y^2 must be negative.
    Therefore, the revised answer is
    x -1 = -(1/12)(y-4)^2 -------------------answer.

    ------------------------------------------------------------
    3. Find the center, vertices, foci and sketch the graph of the ellipse with the following equation. 25x2+12y2=300

    25x^2 +12y^2 = 300 ---------(i)
    Divide both sides by 300 to make the RHS = 1,
    (25/300)x^2 +(12/300)y^2 = 1
    (1/12)x^2 +(1/25) y^2 = 1
    (x^2)/12 +(y^2)/25 = 1
    (x^2)/[(2sqrt(3))^2] +(y^2)/[5^2] = 1 ------(ii)

    That is a "vertical" ellipse in that the major axis is vertical.
    semi-major axis, a = 5 units.
    semi-minor axis, b = 2sqrt(3) units.
    So, focal distance from center, c = sqrt(a^2 -b^2) = sqrt(25 -12) = sqrt(13) units.

    Hence, for the ellipse,
    >>>center is at (0,0).
    >>>vertices are at (0,5) and (0,-5).
    >>>foci are at (0,sqrt(13)) and (0,-sqrt(13)).
    Last edited by ticbol; July 10th 2008 at 06:57 PM. Reason: The horizontal parabola in (2) opens to the left.
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