Isaiah's Math Lessons : Parabola


  • Senior Math Wizard Champion (2011, 2012) and Math Count Champion (2011)


    Parabola Definition:

    The locus of a point that moves it is equidistant from a fixed point called focus and a fixed line called directrix.

    General Equation:

    Ax^2 + Dx + Ey + F = 0 (Axis Parallel to the y –axis)

    Ay^2 + Dx + Ey + F = 0 (Axis Parallel to the x –axis)

    Standard Equation:

    Parabola with vertex @ (h, k)

               (y-k)^2 = 4a (x-h) --- > Opens to the right

               (y-k)^2 = -4a (x-h) -- > Opens to the left

               (x-h)^2 = 4a (y-k) --- > Opens upward

               (x-h)^2 = - 4a (y-k) --- > Opens downward

    KEY FORMULAS:

                4a = Length of Latus Rectum (LR)

                a = focal distance

                (h + a, k) = coordinates of the focus if axis is parallel to the x - axis

                (h, k + a) = coordinates of the focus if axis is parallel to the y - axis

                (h + a, k +/- 2a) = coordinates of the endpoints of LR if axis is parallel to the x – axis

                (h +/- 2a, k + a) = coordinates of the endpoints of LR if axis is parallel to the y - axis

                e = 1 (always constant for parabola)

    Examples and Detailed Solutions:

    1.     Find the general equation of the parabola with focus at (1, 0) and vertex at (2, 0).

    Solution:

    By plotting the points, we can directly say that the parabola is opening to the left or with axis parallel to the x-axis.

    Thus a = 1.

    (y-k)^2 = -4a(x-h)

    y^2 = -4(1)(x-2)

    y^2 = -4(x-2)

    y^2 + 4x - 8 = 0

    2.    Find the general equation of the parabola with vertex at (0, 3) and directrix at x = -1.

    Solution:

    By graphing and plotting the points, this parabola has an axis of symmetry parallel to the y-axis.

    Distance from directrix and vertex is equal to a. Thus, a = 1 and LR = 4. Also, by inspection, the graph opens to the right.

    (y-k)^2 = 4a(x-h)

    (y-3)^2 = 4(1)(x)

    y^2 - 4x - 6y + 9 = 0

    Practice Problems:

    1.     Find the equation of the parabola with axis vertical, vertex at (-1, -1) and passing through  (2, 2).

    2.     What is the equation of the parabola with axis vertical and passing through (0, 0), (1, 0) and (5, -20)?

    Answers:

    1.     x^2 + 2x - 3y - 2 = 0

    2.     x^2 - x + y = 0

     


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