MATH SOLVE

3 months ago

Q:
# What is the equation of the quadratic graph with a focus of (5, −1) and a directrix of y = 1? (1 point)

Accepted Solution

A:

check the picture below.

so the focus point is at 5, -1 and the directrix is above it, meaning is a vertical parabola, and is opening downwards, like in the picture.

keep in mind that the vertex is half-way between those two fellows, at a "p" distance from either, in this case 1 unit, since the parabola is opening downwards, "p" is negative then, or -1, and the vertex will be from 5, -1 up one unit, so 5,0.

[tex]\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}\\\\ -------------------------------\\\\ \begin{cases} h=5\\ k=0\\ p=-1 \end{cases}\implies (x-5)^2=4(-1)(y-0)\implies (x-5)^2=-4y \\\\\\ -\cfrac{1}{4}(x-5)^2=y[/tex]

so the focus point is at 5, -1 and the directrix is above it, meaning is a vertical parabola, and is opening downwards, like in the picture.

keep in mind that the vertex is half-way between those two fellows, at a "p" distance from either, in this case 1 unit, since the parabola is opening downwards, "p" is negative then, or -1, and the vertex will be from 5, -1 up one unit, so 5,0.

[tex]\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}\\\\ -------------------------------\\\\ \begin{cases} h=5\\ k=0\\ p=-1 \end{cases}\implies (x-5)^2=4(-1)(y-0)\implies (x-5)^2=-4y \\\\\\ -\cfrac{1}{4}(x-5)^2=y[/tex]