6.3   The Parabola

A parabola is the set of all points (x, y) in the plane that are equidistant from a fixed line and a fixed point which is not on the fixed line.

In this section, we will discuss the properties of the parabolas. The fixed line is called the directrix of the parabola and the fixed point is the focus. The line through the focus and perpendicular to the directrix is called the axis of the parabola. The point of intersection of the axis and the parabola is called the vertex. A parabola is symmetric about the axis.
fig 6.2-1


Denote the distance between the focus F and the vertex V by p. The vertex is equidistant from the focus and the directrix. Therefore, the distance between focus and directrix is 2p. As illustrated in the figure, the parabola passes through two of the corners of the box that extends from the vertex to the focus along the axis of symmetry and extends 2p units above and 2p units below the axis of symmetry. The line segment passes through the focus and is parallel to the directrix is called the latus rectum. The end points of the latus rectum lie on the curve.
fig 6.2-2


General Form of the Equation of a Parabola

From the general form of the conics equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0, we begin with the case where B = 0. In this case, the term containing xy is not present and it becomes
Ax2 + Cy2 + Dx + Ey + F = 0.
Furthermore, one of A and C equals 0 becomes a parabola. We have already discussed the procedure for identifying the graph of this kind of equation; we complete the squares of the quadratic expressions in x or y, or both. Then we have
Ax2 + Dx + Ey + F = 0
or
Cy2 + Dx + Ey + F = 0.
Above two equations are called the general forms of equation of a parabola. We have learned that the graph of the quadratic function y = ax2 + bx + c, a ≠ 0 is a parabola. If a > 0, the graph opens up and otherwise, it opens down.

Equations of a Parabola: Vertex at (0, 0)

The equation of a parabola is simplest if the coordinate axes are positioned so that the vertex is at the origin and the axis is along the x-axis or y-axis.
We will derive the equation for the parabola. Let P(x, y) be any point on the parabola. Since P is equidistant from the focus F(p, 0) and D(-p, y) is the foot of the perpendicular from P to the directrix, where p > 0, we have PF = PD.
fig 6.2-3

By the distance formula,
$$\small{ PE = \sqrt{(p - x)^2 + (0 - y)^2},}$$ $$\small{ PD = \sqrt{(-p - x)^2 + (y - y)^2}.}$$ Since PF = PD, we have
$$\small{\sqrt{(p - x)^2 + (0 - y)^2} = \sqrt{(-p - x)^2 + (y - y)^2}}$$ p2 - 2px + x2 + y2 = p2 + 2px + x2
y2 = 4px.

The four possible orientations are shown in the following figures. These are called the standard forms of equation of a parabola.
fig 6.2-4

fig 6.2-5

fig 6.2-6

fig 6.2-7

Equations of a parabola:
Vertex at (0, 0) , Focus on Coordinate Axis, p > 0

Equation Vertex Focus Directrix Axis of Symmetry Description
y2 = 4px (0, 0) (p, 0) x = -p horizontal axis, y = 0 opens to the right
y2 = -4px (0, 0) (-p, 0) x = p horizontal axis, y = 0 opens to the left
x2 = 4py (0, 0) (0, p) y = -p vertical axis, x = 0 opens up
x2 = -4py (0, 0) (0, -p) y = p vertical axis, x = 0 opens down

Parabolas can be graphed by using the following four steps :
  • The axis of symmetry is along the x-axis if the equation has a y2 term, and it is along the y-axis if the equation has a x2 term.

  • If the axis of symmetry is along the x-axis, then the parabola
    - opens to the right if the coefficient of x is positive.
    - opens to the left if the coefficient of x is negative.

  • If the axis of symmetry is along the x-axis, the parabola
    - opens up if the coefficient of y is positive.
    - opens down if the coefficient of y is negative.

  • Determine the value of p from the standard equation and draw a box extending p units from the origin along the axis of symmetry in the direction in which the parabola opens and extending 2p units on each side of the axis of symmetry.

  • Using the above facts, sketch the parabola so that its vertex is at the origin and it passes through the corners of the box.

    Example 6.
    Sketch the graphs of the parabolas, showing the vertex, focus, directrix and end points of latus rectum of each equation.
    (a) x2 = 16y     (b) y2 + 8x = 0
    Solution
    (a) x2 = 16y
    Comparing with x2 = 4py, we get p = 4.
    Therefore, the vertex V = (0, 0),   the focus F = (0, 4), the directrix is y = -4 and end points of latus rectum are (-8, 4), (8, 4).
    fig 6.2-8

    (b) y2 + 8x = 0
    y2 = -8x
    Comparing with y2 = -4px, we get   p = 2.
    Therefore, the vertex V = (0, 0),   the focus F(-2, 0), the directrix is x = 2 and end points of latus rectum are (-2, -4), (-2, 4).
    fig 6.2-9


    Example 7.
    Find an equation of the parabola that is symmetric about the y-axis with its vertex at the origin, and passes through (5, 2).
    Solution
    Since the parabola is symmetric about the y-axis, the vertex V = (0, 0) and passes through (5, 2), the equation of the form is
    x2 = 4py
    52 = 4p(2)
    4p = 252.
    Therefore, the equation of the parabola is
    x2 = 252y.

    Exercise 6.2

    In Exercise 1 - 8, (a) sketch the parabola,   (b) show the focus, vertex, directrix and end points of latus rectum.
    1. y2 = 8x
    Solution
    y2 = 8x
    Comparing with y2 = 4px,
    4px = 8x
    p = 8x 4x
    p = 2
    Therefore, the vertex V = (0,0),
    the focus F = (p, 0) = (2, 0),
    the directrix is x = -p = -2,
    end points of latus rectum are
    (p, -2p), (p, 2p) = (2, -4), (2, 4)

    fig 6.2-10

    2. y2 = -6x
    Solution
    y2 = -6x
    Comparing with y2 = -4px,
    -4px = -6x
    p = -6x -4x
    = p = 3 2 = 1.5
    Therefore, the vertex V = (0,0),
    the focus F = (-p, 0) = (-1.5, 0),
    the directrix is x = p = 1.5,
    end points of latus rectum are
    (-p, -2p), (-p, 2p) = (-1.5, 3), (-1.5, -3)
    fig 6.2-11

    3. x + y2 = 0
    Solution
    x + y2 = 0
    y2 = -x
    Comparing with y2 = -4px,
    -4px = -x
    p = -x -4x
    p = 1 4 = 0.25
    Therefore, the vertex V = (0,0),
    the focus F = (-p, 0) = (-0.25, 0),
    the directrix is x = p = 0.25,
    end points of latus rectum are
    (-p, -2p), (-p, 2p) = (-0.25, -0.5), (-0.25, 0.5)
    fig 6.2-12

    4. y = x2
    Solution
    y = x2
    x2 = y
    Comparing with x2 = 4py,
    4py = y
    p = y 4y
    = -1 4
    Therefore, the vertex V = (0,0),
    the focus F = (0, p) = (0, 1 4),
    the directrix is y = -p = - 1 4,
    end points of latus rectum are
    (-2p, p), (2p, p) = (- 1 2, 1 4), (1 2, 1 4)
    fig 6.2-13

    5. y = -x2
    Solution
    y = -x2
    x2 = -y
    Comparing with x2 = -4py,
    -4py = -y
    p = -y -4y
    = 1 4
    Therefore, the vertex V = (0,0),
    the focus F = (0, -p) = (0, - 1 4),
    the directrix is y = p = - 1 4,
    end points of latus rectum are
    (-2p, -p), (2p, -p) = (- 1 2, - 1 4), (1 2, - 1 4)
    fig 6.2-14

    6. y = 3x2
    Solution
    y = 3x2
    x2 = y 3
    Comparing with x2 = 4py,
    4py = y 3
    p = y 3⋅4y
    = 1 12
    Therefore, the vertex V = (0,0),
    the focus F = (0, p) = (0, 1 12),
    the directrix is y = -p = - 1 12,
    end points of latus rectum are
    (-2p, p), (2p, p) = (- 1 6, 1 12), (1 6, 1 12)
    fig 6.2-15

    7. y = -3x2
    Solution
    y = -3x2
    x2 = - y 3
    Comparing with x2 = -4py,
    -4py = - y 3
    p = -y 3⋅(-4)y
    = 1 12
    Therefore, the vertex V = (0,0),
    the focus F = (0, -p) = (0, - 1 12),
    the directrix is y = p = 1 12,
    end points of latus rectum are
    (-2p, -p), (2p, -p) = (- 1 6, - 1 12), (1 6, - 1 12)
    fig 6.2-16

    8. x2 + 6y = 0
    Solution
    x2 = -6y
    Comparing with x2 = -4py,
    -4py = - 6y
    p = -6y -4y
    = 3 2
    Therefore, the vertex V = (0,0),
    the focus F = (0, -p) = (0, - 3 2),
    the directrix is y = p = 3 2,
    end points of latus rectum are
    (-2p, -p), (2p, -p) = (-3 , - 3 2), (3, - 3 2)
    fig 6.2-17


    In Exercise 9 - 16, find an equation for the parabola satisfying the given conditions.
    9. Vertex (0, 0),   focus (3, 0).
    Solution
    Vertex (0,0),
    focus (3, 0) = (p, 0)
    p = 3
    The equation of the parabola is
    y2 = 4px
    = 4(3)x
    = 12x

    10. Vertex (0, 0),   focus (0, -4)
    Solution
    Vertex (0, 0)
    focus (0, -4) = (0, -p)
    -p = -4
    p = 4
    The equation of the parabola is
    x2 = -4py
    = -4 (4)y
    = -16y

    11. Vertex (0, 0).   directrix x = 7.
    Solution
    Vertex (0, 0)
    directrix x = 7
    p = 7
    The equation of the parabola is
    y2 = -4px
    = -4 (7)x
    = -28x

    12. Vertex (0, 0),   directrix y = 1 2.
    Solution
    Vertex (0, 0)
    directrix y = 1 2.
    p = 1 2.
    The equation of the parabola is
    x2 = -4py
    = -4 (1 2)y
    = -2y

    13. Focus (0, -3),   directrix y = 3.
    Solution
    Focus (0, -3),
    directrix y = 3
    p = 3
    The equation of the parabola is
    x2 = -4py
    = (-4)⋅3 y
    = -12y

    14. Focus (6, 0),   directrix x = -6
    Solution
    Focus (6, 0) = (p, 0)
    p = 6
    directrix x = -6
    x = -p
    -p = -6
    p = 6
    The equation of the parabola is
    y2 = 4px
    = 4⋅6x
    = 24x

    15. Vertex (0, 0);   symmetric about the x-axis; passes through (2, 2).
    Solution
    Since the parabola is symmetric about the x-axis, Vertex V = (0, 0) and passes through (2, 2), the equation of the form is
    y2 = 4px
    At (2, 2)
    22 = 4p(2)
    4 2 = 4p
    4p = 2
    Threrfore, the equation of the parabola is
    y2 = 4px
    = 2x       (∵ 4p = 2)

    16. Vertex (0, 0);   symmetric about the y-axis; passes through (-1, 3).
    Solution
    Vertex V = (0, 0)
    x2 = 4py
    At (-1, 3)
    (-1)2 = 4p(3)
    1 3 = 4p
    Threrfore, the equation of the parabola is
    x2 = 4py
    x2 = 1 3y       (∵ 4p = 1 3)
    3x2 = y

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