Parabola

BigGlenntheHeavy

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Find an equation of the following parabola.

Directrix y = 1, length of latus rectum = 8, opens downward.
 
Hello, BigGlenntheHeavy!

There is insufficient information for a unique solution.


Find an equation of the following parabola:
. . Directrix: y=1\displaystyle y = 1, length of latus rectum = 8, opens downward.

Since the parabola opens downward,\displaystyle \text{Since the parabola opens downward,}

\(\displaystyle \text{the general form is: }\:(x-h)^2 \:=\:-4p(y-k)\)

. . where (h,k) is the Vertex, and p is the distance from the Vertex to the Focus\displaystyle \text{where }(h,k)\text{ is the Vertex, and }p\text{ is the distance from the Vertex to the Focus}
. . and the distance from the Vertex to the Directrix.\displaystyle and\text{ the distance from the Vertex to the Directrix.}


Fact: the length of the Latus Rectum is 4p.\displaystyle \text{Fact: the length of the Latus Rectum is }4p.

. . Hence: 4p=8p=2\displaystyle \text{Hence: }\:4p \,=\,8 \quad\Rightarrow\quad p \,=\,2


So far, we have:


Code:
      |
    . + . . . . . . . y = 1
      |
      |
  ----+----------------------
      |
      |     (h,-1)
      |       o
      |   *       *
      | *           *
      |*             *
      |
      *               *
      |

The directrix is: y=1\displaystyle \text{The directrix is: }\,y = 1

\(\displaystyle \text{We have: }\:p = 2.\)

\(\displaystyle \text{The vertex is: }\:(h,-1)\)

\(\displaystyle \text{The equation is: }\:(x-h)^2 \:=\:-8(y+1)\)

 
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