Question

Write the equation for a parabola with a focus at (0,-5)(0,−5)left parenthesis, 0, comma, minus, 5, right parenthesis and a directrix at y=-3

Answer 1

Now, notice, the focus point is at 0, -5, down below the x-axis, whilst the directrix is above it, meaning is a vertical parabola, and is opening downwards.  That means the squared variable is the "x".

now, keeping in mind that, the vertex is half-way between these two fellows, from y = -5 and y = -3, there are 2 units, and the vertex is half-way there.  Check the picture below.

now, notice the "p" distance from the vertex to either fellow, is 1 unit, however, because the parabola is opening downwards, is negative 1.

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

$\bf \begin{cases} h=0\\ k=-4\\ p=-1 \end{cases}\implies (x-0)^2=4(-1)[y-(-4)] \\\\\\ x^2=-4(y+4)\implies -\cfrac{1}{4}x^2=y+4\implies -\cfrac{1}{4}x^2-4=y$