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Conics - Parabola

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Examples
Lessons
  1. vertical parabola VS. horizontal parabola
    Sketch the following vertical parabolas:
    i) y=x2y = {x^2}
    ii) y=2x2y = 2{x^2}
    iii) y=2(x+3)2+1y = 2{\left( {x + 3} \right)^2} + 1
    1. Sketch the following horizontal parabolas:
      i) x=y2x = {y^2}
      ii) x=12y2x = \frac{1}{2}{y^2}
      iii) x=12(y−1)2−3x = \frac{1}{2}{\left( {y - 1} \right)^2} - 3
      1. converting quadratic functions to vertex form by "completing the square"
        Convert each quadratic function from general form to vertex form by completing the square.
        1. y=2x2−12x+10y = 2{x^2} - 12x + 10
        2. y2−10y−4x+13=0{y^2} - 10y - 4x + 13 = 0
      2. finding the focus and directrix using the formula: p=14ap = \frac{1}{{4a}}
         For each quadratic function, state the:
        i) vertex
        ii) axis of symmetry
        iii) focus
        iv) directrix
        1. y=18(x−6)2+3y = \frac{1}{8}{\left( {x - 6} \right)^2} + 3
        2. −12(x+1)=(y+4)2 - 12\left( {x + 1} \right) = {\left( {y + 4} \right)^2}
        3. y2−10y−4x+13=0{y^2} - 10y - 4x + 13 = 0
      Topic Notes
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      parabola: a curve formed from all the points that are equidistant from the focus and the directrix.
      vertex: midway between the focus and the directrix
      focus: a point inside the parabola
      directrix: a line outside the parabola and perpendicular to the axis of symmetry

      conics formula for parabola:
      p=14ap = \frac{1}{{4a}} p: distance between the vertex and the focus / directrix.
      a: coefficient of the squared term
      Basic Concepts
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