{"id":1041,"date":"2025-06-20T17:30:18","date_gmt":"2025-06-20T17:30:18","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=1041"},"modified":"2025-09-10T18:36:10","modified_gmt":"2025-09-10T18:36:10","slug":"conic-sections-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/conic-sections-learn-it-2\/","title":{"raw":"Conic Sections: Learn It 2","rendered":"Conic Sections: Learn It 2"},"content":{"raw":"

Parabolas<\/h2>\r\n

A parabola forms when a plane intersects a cone parallel to the generating line. Unlike other conic sections, the plane intersects only one nappe of the cone.<\/p>\r\n\r\n

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parabola<\/h3>\r\nA parabola is the set of all whose distance from a fixed point (the focus<\/strong>) equals the distance from a fixed line (the directrix<\/strong>).\r\n\r\n<\/section>The vertex<\/strong> of a parabola is the point halfway between the focus and the directrix. This point represents the parabola's turning point.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"417\"]\"A Figure 3. A typical parabola in which the distance from the focus to the vertex is represented by the variable [latex]p[\/latex].[\/caption]\r\n

Using the distance definition and the distance formula, we can derive the equation of a parabola.<\/p>\r\n\r\n

Distance Formula<\/strong>: For points [latex]P(x_1, y_1)[\/latex] and [latex]Q(x_2, y_2)[\/latex], the distance is [latex]d(P,Q) = \\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[\/latex].<\/section>From Figure 3, we apply the parabola definition: the distance from focus [latex]F[\/latex] to point [latex]P[\/latex] equals the distance from [latex]P[\/latex] to point [latex]Q[\/latex] on the directrix.\r\n
[latex]\\begin{array}{ccc}\\hfill d\\left(F,P\\right)& =\\hfill & d\\left(P,Q\\right)\\hfill \\\\ \\hfill \\sqrt{{\\left(0-x\\right)}^{2}+{\\left(p-y\\right)}^{2}}& =\\hfill & \\sqrt{{\\left(x-x\\right)}^{2}+{\\left(-p-y\\right)}^{2}}.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

Squaring both sides and simplifying:<\/p>\r\n\r\n

[latex]\\begin{array}{ccc}\\hfill {x}^{2}+{\\left(p-y\\right)}^{2}& =\\hfill & {0}^{2}+{\\left(-p-y\\right)}^{2}\\hfill \\\\ \\hfill {x}^{2}+{p}^{2}-2py+{y}^{2}& =\\hfill & {p}^{2}+2py+{y}^{2}\\hfill \\\\ \\hfill {x}^{2}-2py& =\\hfill & 2py\\hfill \\\\ \\hfill {x}^{2}& =\\hfill & 4py.\\hfill \\end{array}[\/latex]<\/div>\r\n
Recall: Transformations of graphs\r\n[latex]\\\\[\/latex]\r\n<\/strong>For function [latex]y = f(x)[\/latex], the graph [latex]y = f(x - h) + k[\/latex] shifts vertically by [latex]k[\/latex] units and horizontally by [latex]h[\/latex] units.\r\n
    \r\n \t
  • Positive [latex]k[\/latex]: shift up; negative [latex]k[\/latex]: shift down<\/li>\r\n \t
  • Positive [latex]h[\/latex]: shift right; negative [latex]h[\/latex]: shift left<\/li>\r\n<\/ul>\r\nThe equation [latex]y = f(x - h) + k[\/latex] is equivalent to [latex]y - k = f(x - h)[\/latex]. When you replace [latex]y[\/latex] with [latex]y - k[\/latex] and [latex]x[\/latex] with [latex]x - h[\/latex] in any equation, the graph shifts according to these rules.\r\n\r\n<\/section>What if the vertex isn't at the origin? We use coordinates\u00a0 [latex]\\left(h,k\\right)[\/latex] to represent the vertex location. When the focus sits directly above the vertex, it has coordinates [latex](h, k+p)[\/latex] and the directrix becomes the line [latex]y = k - p[\/latex].\r\n\r\nUsing the same derivation process gives us [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex]. Solving for [latex]y[\/latex] leads to our key theorem.\r\n\r\n
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    standard form of a parabola<\/h3>\r\n

    Given a parabola opening upward with vertex located at [latex]\\left(h,k\\right)[\/latex] and focus located at [latex]\\left(h,k+p\\right)[\/latex]], the equation is:<\/p>\r\n\r\n

    [latex]y=\\frac{1}{4p}{\\left(x-h\\right)}^{2}+k[\/latex].<\/div>\r\n \r\n

    This is the standard form<\/strong> of a parabola.<\/p>\r\n\r\n<\/section>\r\n

    Parabolas can open in four directions: up, down, left, or right. Each orientation has its own standard form equation, as shown in Figure 4.<\/p>\r\n\r\n

    <\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"800\"]\"This Figure 4. Four parabolas, opening in various directions, along with their equations in standard form.[\/caption]<\/figure>\r\n

    You'll also encounter parabolas written in general form<\/strong>, where the values of [latex]h[\/latex], [latex]k[\/latex], and [latex]p[\/latex] aren't immediately visible:<\/p>\r\n\r\n

      \r\n \t
    • Vertical parabolas (opens up\/down): [latex]ax^2 + bx + cy + d = 0[\/latex]<\/li>\r\n \t
    • Horizontal parabolas (opens left\/right): [latex]ay^2 + bx + cy + d = 0[\/latex]<\/li>\r\n<\/ul>\r\n
      Converting to Standard Form<\/strong>: Use completing the square to convert from general form to standard form. This reveals the vertex coordinates and makes graphing much easier.<\/section>
      \r\n
      \r\n

      Put the equation [latex]{x}^{2}-4x - 8y+12=0[\/latex] into standard form and graph the resulting parabola.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n

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      Since y<\/em> is not squared in this equation, we know that the parabola opens either upward or downward. Therefore we need to solve this equation for y,<\/em> which will put the equation into standard form. To do that, first add [latex]8y[\/latex] to both sides of the equation:<\/p>\r\n\r\n

      [latex]8y={x}^{2}-4x+12[\/latex].<\/div>\r\n \r\n

      The next step is to complete the square on the right-hand side. Start by grouping the first two terms on the right-hand side using parentheses:<\/p>\r\n\r\n

      [latex]8y=\\left({x}^{2}-4x\\right)+12[\/latex].<\/div>\r\n \r\n

      Next determine the constant that, when added inside the parentheses, makes the quantity inside the parentheses a perfect square trinomial. To do this, take half the coefficient of x<\/em> and square it. This gives [latex]{\\left(\\frac{-4}{2}\\right)}^{2}=4[\/latex]. Add 4 inside the parentheses and subtract 4 outside the parentheses, so the value of the equation is not changed:<\/p>\r\n\r\n

      [latex]8y=\\left({x}^{2}-4x+4\\right)+12 - 4[\/latex].<\/div>\r\n \r\n

      Now combine like terms and factor the quantity inside the parentheses:<\/p>\r\n\r\n

      [latex]8y={\\left(x - 2\\right)}^{2}+8[\/latex].<\/div>\r\n

      Finally, divide by 8:<\/p>\r\n\r\n

      [latex]y=\\frac{1}{8}{\\left(x - 2\\right)}^{2}+1[\/latex].<\/div>\r\n \r\n

      This equation is now in standard form. Comparing this to equations for parabolas gives [latex]h=2[\/latex], [latex]k=1[\/latex], and [latex]p=2[\/latex]. The parabola opens up, with vertex at [latex]\\left(2,1\\right)[\/latex], focus at [latex]\\left(2,3\\right)[\/latex], and directrix [latex]y=-1[\/latex]. The graph of this parabola appears as follows.<\/p>\r\n\r\n

      <\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"417\"]\"A Figure 5. The parabola in [link].[\/caption]<\/figure>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
      Watch the following video to see the worked solution to the example above.