Question

Given a focus of (4, 5) and directrix of y= -3 , find the equation of the parabola.

Answer 1

Check the picture below.

notice, the focus point is at 4,5 whilst the directrix line is at y = -3, below the focus point, meaning the parabola is vertical and opening upwards.

keeping in mind that the vertex is "p" distance from either of these fellows, then the vertex is half-way between both of them, notice in the picture, the distance from y = 5 to y = -3 is 8 units, half that is 4 units, thus the vertex 4 units from the focus or 4 units from the directrix, that puts it at (4,1), whilst "p" is 4, since the parabola is opening upwards, is a positive 4 then.

$\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} (y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\ \boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}})} \end{array} \qquad \begin{array}{llll} vertex\ ({{ h}},{{ k}})\\\\ {{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -------------------------------$

$\bf \begin{cases} h=4\\ k=1\\ p=4 \end{cases}\implies (x-4)^2=4(4)(y-1)\implies (x-4)^2=16(y-1) \\\\\\ \cfrac{1}{16}(x-4)^2=y-1\implies \cfrac{1}{16}(x-4)^2+1=y$