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Write an equation of the parabola with focus F(0,-3) and directrix y=3

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check the picture below, so the parabola looks more or less like so.


bearing in mind that the vertex is half-way between the directrix and the focus point, so that puts it in the origin.

the distance from the vertex to the directrix or focus point is "p" units, in this case, just 3 units, however, the parabola is opening downwards, meaning "p" is negative, so p = -3.



\bf \textit{parabola vertex form with focus point distance} \\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \boxed{4p(y- k)=(x- h)^2} \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=0\\ k=0\\ p=-3 \end{cases}\implies 4(-3)(y-0)=(x-0)^2 \\\\\\ -12y=x^2\implies \blacktriangleright y=-\cfrac{1}{12}x^2\blacktriangleleft

Write an equation of the parabola with focus F(0,-3) and directrix y=3-example-1
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User Sonhu
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