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Determine the elements of the parabola and graph, whose general equation is: x2 - 12x + 10y - 4= 0

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Answer to a math question Determine the elements of the parabola and graph, whose general equation is: x2 - 12x + 10y - 4= 0

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Esmeralda
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59 Answers
Para determinar los elementos de la parábola con la ecuación general x^2 - 12x + 10y - 4 = 0, primero necesitamos convertirla a su forma estándar completa. La ecuación general de una parábola es de la forma Ax^2 + Bx + Cy + D = 0.

Dada la ecuación x^2 - 12x + 10y - 4 = 0, vamos a completar el cuadrado para x y despejar y:

1. Pasamos los términos con x a un lado y los términos independientes al otro lado de la ecuación:
x^2 - 12x + 10y = 4

2. Completamos el cuadrado para x:
(x^2 - 12x + 36) + 10y = 4 + 36
(x - 6)^2 + 10y = 40

3. Despejamos y:
10y = - (x - 6)^2 + 40
y = -\frac{1}{10}(x - 6)^2 + 4

Ahora que hemos convertido la ecuación general a la forma estándar completa, podemos identificar los elementos de la parábola:

- Vértice: (h, k) = (6, 4)
- Eje de simetría: x = 6
- Distancia focal: |p| = \frac{1}{4a} = \frac{1}{4(1/10)} = 2.5
- Foco: (h, k + p) = (6, 4 + 2.5) = (6, 6.5)

Graficando la parábola:
\begin{array}{ c }\begin{tikzpicture} \begin{axis}[ axis lines = center, xlabel = x, ylabel = y, xmin = 0, xmax = 12, ymin = 0, ymax = 10, ] \addplot [ domain=0:12, samples=100, color=blue, ] {-1/10*((x-6)^2)+4}; \addplot [only marks, mark=*] coordinates {(6,4) (6,6.5)}; \end{axis}\end{tikzpicture}\end{array}

\textbf{Respuesta: } La parábola tiene vértice en el punto (6, 4), eje de simetría en x = 6, una distancia focal de 2.5 y su foco se encuentra en el punto (6, 6.5).

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