Write the equation for a parabola with a focus

Ответ:

castilloitc1120

12.03.2022

Now, notice, the focus point is at 0, -5, down below the x-axis, whilst the directrix is above it, meaning is a vertical parabola, and is opening downwards. That means the squared variable is the "x".

now, keeping in mind that, the vertex is half-way between these two fellows, from y = -5 and y = -3, there are 2 units, and the vertex is half-way there. Check the picture below.

now, notice the "p" distance from the vertex to either fellow, is 1 unit, however, because the parabola is opening downwards, is negative 1.

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

$\bf \begin{cases}
h=0\\
k=-4\\
p=-1
\end{cases}\implies (x-0)^2=4(-1)[y-(-4)]
\\\\\\
x^2=-4(y+4)\implies -\cfrac{1}{4}x^2=y+4\implies -\cfrac{1}{4}x^2-4=y$

Write the equation for a parabola with a focus at (0,-5)(0,−5)left parenthesis, 0, comma, minu

Write the equation for a parabola with a focus at (0,-5)(0,−5)left parenthesis, 0, comma, minu

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Write the equation for a parabola with a focus at (0,-5)(0,−5)left parenthesis, 0, comma, minus, 5, right parenthesis and a directrix at y=-3

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