What is the equation of the quadratic graph with a focus of (1 ,3) and a directrix of y=1?

Mathematics

1 answer:

zimovet [89]2 years ago

4 0

Check the picture below.

now, we know the directrix is at y = 1, and the focus point is at 1,3, well, notice the picture, the distance between those fellows is just 2 units.

the vertex is half-way between those fellows, therefore, the vertex will be at 1,2.

the distance "p", from the vertex to either the directrix or focus, is really just 1 unit. Since the focus point is above the directrix, is a vertical parabola, and it opens upwards, like in the picture, and since it opens up, the "p" value is positive, or +1.

$\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
4p(x- h)=(y- k)^2
\\\\
\boxed{4p(y- k)=(x- h)^2}
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ( h, k)\\\\
 p=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------\\\\
\begin{cases}
h=1\\
k=2\\
p=1
\end{cases}\implies 4(1)(y-2)=(x-1)^2\implies 4(y-2)=(x-1)^2
\\\\\\
y-2=\cfrac{1}{4}(x-1)^2\implies y=\cfrac{1}{4}(x-1)^2+2$

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