What is the equation of a parabola with (−2, 4) as its focus and y = 6 as its directrix? Enter the equation in the box.

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notice the picture below the directrix is above the focus point, meaning the parabola is vertical and is opening downwards now, "p" is the distance from the vertex to the focus point or the directrix, so that means, the vertex is between those two fellows, over the axis of symmetry, x = -2, since "p" is 1 unit, that puts the vertex at -2,5 since the parabola is opening downwards, that means the "p" value is negative, so is -1 [latex]\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} (y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\ \boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}}) }\\ \end{array} \qquad \begin{array}{llll} vertex\ ({{ h}},{{ k}})\\\\ {{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -----------------------------\\\\[/latex] [latex]\bf \begin{cases} p=-1\\ h=-2\\ k=5 \end{cases}\implies (x-(-2))^2=4(-1)(y-5) \\\\\\ (x+2)^2=-4(y-5)\implies -\cfrac{1}{4}(x+2)^2=y-5 \\\\\\ \boxed{-\cfrac{1}{4}(x+2)^2+5=y}[/latex]

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