Using a directrix of y = 2 and a focus of (3, −4), what quadratic function is created?

Asked by maham237 @ 18/12/2020 in Mathematics viewed by 320 People

f(x) = one twelfth (x − 3)2 − 1

f(x) = − one twelfth (x − 3)2 − 1

f(x) = − one sixth (x + 3)2 + 1

f(x) = one sixth (x − 3)2 + 1

Answered by maham237 @ 18/12/2020

Check the picture below, so keeping in mind that, the distance "p" from the vertex to either the focus point or directrix is the same, that means the vertex is half-way in between those two folks.

now, the directrix is above, and the focus point is below, that means is a vertical parabola, meaning the "x" is the squared variable, and it opens downwards, meaning the distance "p" is a negative value, in this case is -3.

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

$\bf \begin{cases}
h=3\\
k=-1\\
p=-3
\end{cases}\implies (x-3)^2=4(-3)[y-(-1)]
\\\\\\
(x-3)^2=-12(y+1)\implies \cfrac{(x-3)^2}{-12}=y+1
\\\\\\
-\cfrac{1}{12}(x-3)^2-1=y$

Using a directrix of y = 2 and a focus of (3, −4), what quadratic function is created?

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