MATH SOLVE

2 months ago

Q:
# Find the standard form of the equation of the parabola with a focus at (5, 0) and a directrix at x = -5.

Accepted Solution

A:

let's notice, the focus is at (5,0) and the directrix at x = -5.

keep in mind that there's a distance "p" from the vertex to either of those fellows, therefore, if the focus is at 5,0 and the directrix x = -5, the vertex is half-way between them, check the picture below.

notice the distance "p" there, now, is a horizontal parabola, opening to the right, meaning the value for "p" is positive, or just 5, thus

[tex]\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} \boxed{(y-{{ k}})^2=4{{ p}}(x-{{ h}})} \\\\ (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=0\\ k=0\\ p=5 \end{cases}\implies (y-0)^2=4(5)(x-0)\implies y^2=20x \\\\\\ \cfrac{y^2}{20}=x\implies \cfrac{1}{20}y^2=x[/tex]

keep in mind that there's a distance "p" from the vertex to either of those fellows, therefore, if the focus is at 5,0 and the directrix x = -5, the vertex is half-way between them, check the picture below.

notice the distance "p" there, now, is a horizontal parabola, opening to the right, meaning the value for "p" is positive, or just 5, thus

[tex]\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} \boxed{(y-{{ k}})^2=4{{ p}}(x-{{ h}})} \\\\ (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=0\\ k=0\\ p=5 \end{cases}\implies (y-0)^2=4(5)(x-0)\implies y^2=20x \\\\\\ \cfrac{y^2}{20}=x\implies \cfrac{1}{20}y^2=x[/tex]