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NNADVOKAT [17]
3 years ago
14

(03.09 MC) Derive the equation of the parabola with a focus at (−2, 4) and a directrix of y = 6. Put the equation in standard fo

rm

Mathematics
2 answers:
Serjik [45]3 years ago
7 0

Answer: f(x) = −one fourth x2 − x + 4


Nookie1986 [14]3 years ago
5 0
Check the picture below.

so the focus point is at -2, 4 and the directrix is at y = 6, now, keeping in mind that the vertex is half-way between those two fellows, from 4 to 6, it'd be the y-coordinate of 5, and therefore, the vertex is at -2,5, as you see there in the picture, and the parabola looks like so.  Since the parabola is a vertical one, the squared variable is the "x".

notice the distance "p", is just 1 unit, however, since the parabola is opening downwards, "p" is negative, and thus -1.

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

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