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tester [92]
3 years ago
14

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

Mathematics
1 answer:
Elena-2011 [213]3 years ago
8 0
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

\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

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