Question

Derive the equation of the parabola with a focus at (−5, 5) and a directrix of y = −1.

Answer 1

Answer: D

Step-by-step explanation: took the test

Answer 2

check the picture below.

so, the left-part of the picture is the focus point and the directrix... .so.. notice, the focus is above the directrix, meaning is a vertical parabola and is opening upwards.

now, looking at the right-part of the picture, bear in mind that, the vertex is "p" units away from the directrix and the focus, so, the vertex is half-way between those fellows, notice in the picture what "p" is, keep in mind that, because the parabola is opening upwards, "p" is a positive unit, thus is 3.


`\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=-5\\ k=2\\ p=3 \end{cases}\implies [x-(-5)]^2=4(3)(y-2) \\\\\\ (x+5)^2=12(y-2)\implies \cfrac{(x+5)^2}{12}=y-2\implies \cfrac{(x+5)^2}{12}+2=y \\\\\\ \cfrac{1}{12}(x+5)^2+2=y`