text: # What is the standard form of the equation of the parabola with a directrix at x=9 and a focus at (8,4)?
  
  Aug 26, 2017
  
  The standard form is: $x = - \frac{1}{2} {y}^{2} + 4 y + \frac{1}{2}$
  
  #### Explanation:
  
  Because the directrix is a vertical line, one knows that the vertex form of the equation for the parabola is:
  
  $x = \frac{1}{4 f} {\left(y - k\right)}^{2} + h \text{ [1]}$
  
  where $\left(h , k\right)$ is the vertex and $f$ is the signed horizontal distance from the vertex to the focus.
  
  The x coordinate of the vertex halfway between the directrix and the focus :
  
  $h = \frac{9 + 8}{2}$
  
  $h = \frac{17}{2}$
  
  Substitute into equation [1]:
  
  $x = \frac{1}{4 f} {\left(y - k\right)}^{2} + \frac{17}{2} \text{ [2]}$
  
  The y coordinate of the vertex is the same as the y coordinate of the focus:
  
  $k = 4$
  
  Substitute into equation [2]:
  
  $x = \frac{1}{4 f} {\left(y - 4\right)}^{2} + \frac{17}{2} \text{ [3]}$
  
  The value of $f$ is the signed horizontal distance from the vertex to the focus#
  
  $f = 8 - \frac{17}{2}$
  
  $f = - \frac{1}{2}$
  
  Substitute into equation [3]:
  
  $x = \frac{1}{4 \left(- \frac{1}{2}\right)} {\left(y - 4\right)}^{2} + \frac{17}{2}$
  
  This is the vertex form:
  
  $x = - \frac{1}{2} {\left(y - 4\right)}^{2} + \frac{17}{2}$
  
  Expand the square:
  
  $x = - \frac{1}{2} \left({y}^{2} - 8 y + 16\right) + \frac{17}{2}$
  
  Use the distributive property:
  
  $x = - \frac{1}{2} {y}^{2} + 4 y - 8 + \frac{17}{2}$
  
  Combine like terms:
  
  $x = - \frac{1}{2} {y}^{2} + 4 y + \frac{1}{2}$
  
  Here is a graph of the standard form, the focus, the vertex, and the directrix:
explanation: The document contains a math question and a detailed answer. The question asks for the standard form of the equation of a parabola given its directrix and focus. The answer provides a step-by-step derivation of the equation.
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distilabel_metadata.content: # What is the standard form of the equation of the parabola with a directrix at x=9 and a focus at (8,4)?
  
  Aug 26, 2017
  
  The standard form is: $x = - \frac{1}{2} {y}^{2} + 4 y + \frac{1}{2}$
  
  #### Explanation:
  
  Because the directrix is a vertical line, one knows that the vertex form of the equation for the parabola is:
  
  $x = \frac{1}{4 f} {\left(y - k\right)}^{2} + h \text{ [1]}$
  
  where $\left(h , k\right)$ is the vertex and $f$ is the signed horizontal distance from the vertex to the focus.
  
  The x coordinate of the vertex halfway between the directrix and the focus :
  
  $h = \frac{9 + 8}{2}$
  
  $h = \frac{17}{2}$
  
  Substitute into equation [1]:
  
  $x = \frac{1}{4 f} {\left(y - k\right)}^{2} + \frac{17}{2} \text{ [2]}$
  
  The y coordinate of the vertex is the same as the y coordinate of the focus:
  
  $k = 4$
  
  Substitute into equation [2]:
  
  $x = \frac{1}{4 f} {\left(y - 4\right)}^{2} + \frac{17}{2} \text{ [3]}$
  
  The value of $f$ is the signed horizontal distance from the vertex to the focus#
  
  $f = 8 - \frac{17}{2}$
  
  $f = - \frac{1}{2}$
  
  Substitute into equation [3]:
  
  $x = \frac{1}{4 \left(- \frac{1}{2}\right)} {\left(y - 4\right)}^{2} + \frac{17}{2}$
  
  This is the vertex form:
  
  $x = - \frac{1}{2} {\left(y - 4\right)}^{2} + \frac{17}{2}$
  
  Expand the square:
  
  $x = - \frac{1}{2} \left({y}^{2} - 8 y + 16\right) + \frac{17}{2}$
  
  Use the distributive property:
  
  $x = - \frac{1}{2} {y}^{2} + 4 y - 8 + \frac{17}{2}$
  
  Combine like terms:
  
  $x = - \frac{1}{2} {y}^{2} + 4 y + \frac{1}{2}$
  
  Here is a graph of the standard form, the focus, the vertex, and the directrix:
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and a detailed answer. The question asks for the standard form of the equation of a parabola given its directrix and focus. The answer provides a step-by-step derivation of the equation.
  
  Label: question-and-answer