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Conics: Parabolas:
  Finding Information from the Equation
(page 2 of 4)

Sections: Introduction, Finding information from the equation, Finding the equation from information, Word problems & Calculators


  • Graph x2 = 4y and state the vertex, focus, axis of symmetry, and directrix.
  • This is the same graphing that I've done in the past: y = (1/4)x2. So I'll do the graph as usual:

x = -4, y = (1/4)(-4)^2 = 4; x = -2, y = (1/4)(-2)^2 = 1; x = 0, y = (1/4)(0)^2 = 0; x = 2, y = (1/4)(2)^2 = 1; x = 4, y = (1/4)(4)^2 = 4

 

graph of y = (1/4) x^2

    The vertex is obviously at the origin, but I need to "show" this "algebraically" by rearranging the given equation into the conics form:

      x2 = 4y Copyright Elizabeth Stapel 2010-2011 All Rights Reserved
      (x 0)2 = 4(y 0)

    This rearrangement "shows" that the vertex is at (h, k) = (0, 0). The axis of symmetry is the vertical line right through the vertex: x = 0. (I can always check my graph, if I'm not sure about this.) The focus is "p" units from the vertex. Since the focus is "inside" the parabola and since this is a "right side up" graph, the focus has to be above the vertex.

    From the conics form of the equation, shown above, I look at what's multiplied on the unsquared part and see that 4p = 4, so p = 1. Then the focus is one unit above the vertex, at (0, 1), and the directrix is the horizontal line y = 1, one unit below the vertex.

      vertex: (0, 0); focus: (0, 1); axis of symmetry: x = 0; directrix: y = 1

  • Graph y2 + 10y + x + 25 = 0, and state the vertex, focus, axis of symmetry, and directrix.
  • Since the y is squared in this equation, rather than the x, then this is a "sideways" parabola. To graph, I'll do my T-chart backwards, picking y-values first and then finding the corresponding x-values for x = y2 10y 25:

      chart of values

      graph of x = -y^2 - 10y - 25

    To convert the equation into conics form and find the exact vertex, etc, I'll need to convert the equation to perfect-square form. In this case, the squared side is already a perfect square, so:

      y2 + 10y + 25 = x
      (y + 5)2 = 1(x 0)

    This tells me that 4p = 1, so p = 1/4. Since the parabola opens to the left, then the focus is 1/4 units to the left of the vertex. I can see from the equation above that the vertex is at (h, k) = (0, 5), so then the focus must be at (1/4, 5). The parabola is sideways, so the axis of symmetry is, too. The directrix, being perpendicular to the axis of symmetry, is then vertical, and is 1/4 units to the right of the vertex. Putting this all together, I get:

      vertex: (0, 5); focus: (1/4, 5); axis of symmetry: y = 5; directrix: x = 1/4

  • Find the vertex and focus of y2 + 6y + 12x 15 = 0

    The y part is squared, so this is a sideways parabola. I'll get the y stuff by itself on one side of the equation, and then complete the square to convert this to conics form.

      y2 + 6y 15 = 12x
      y
      2 + 6y + 9 15 = 12x + 9

      (y + 3)2 15 = 12x + 9

      (y + 3)2 = 12x + 9 + 15 = 12x + 24

      (y + 3)2 = 12(x 2)

      (y (3))2 = 4(3)(x 2)

    Then the vertex is at (h, k) = (2, 3) and the value of p is 3. Since y is squared and p is negative, then this is a sideways parabola that opens to the left. This puts the focus 3 units to the left of the vertex.

      vertex: (2, 3); focus: (1, 3)

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Cite this article as:

Stapel, Elizabeth. "Conics: Parabolas: Finding Information From the Equation." Purplemath.
    Available from
http://www.purplemath.com/modules/parabola2.htm.
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