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raketka [301]
2 years ago
15

Ecuación de la hipérbola con centro en (0;0), focos en abrir paréntesis 0 coma espacio menos raíz cuadrada de 28 cerrar paréntes

is espacio y espacio abrir paréntesis 0 coma espacio raíz cuadrada de 28 cerrar paréntesis espacio ,eje conjugado = 2 raíz cuadrada de 3
Mathematics
1 answer:
yaroslaw [1]2 years ago
7 0

Answer:

\frac{y^{2}}{25}-\frac{x^{2}}{3}=1

Step-by-step explanation:

Para resolver este problema debemos tomar en cuenta los datos que nos dan y la ecuación de una hipérbola. Comencemos con los datos:

centro: (0,0)

focos: (0,-\sqrt{28}),(0,\sqrt{28})

eje conjugado = 2\sqrt{3}

por los focos podemos ver que la hipérbola se dirige hacia el eje y, por lo que debemos tomar la siguiente forma de la ecuación de la parábola:

\frac{y^{2}}{a^{2}}+\frac{x^{2}}{b^{2}}=1

de los focos podemos obtener que:

c=\sqrt{28}

y del eje conjugado podemos saber que al dividir la longitud del eje conjugado dentro de 2 obtenemos b, así que:

b=\sqrt{3}

podemos utilizar la siguiente fórmula para obtener a:

c^{2}-a^{2}=b^{2}

si despejamos a en la ecuación obtenemos lo siguiente:

a=\sqrt{c^{2}-b^{2}}

ahora podemos sustituir los valores:

a=\sqrt{(\sqrt{28})^{2}-(\sqrt{3})^{2}}

a=\sqrt{28-3}

a=\sqrt{25}

a=5

así que media vez conozcamos a, podemos sustituir los datos en la ecuación de la hipérbola así que obtenemos lo siguiente:

\frac{y^{2}}{a^{2}}+\frac{x^{2}}{b^{2}}=1

\frac{y^{2}}{(5)^{2}}+\frac{x^{2}}{(\sqrt{3})^{2}}=1

\frac{y^{2}}{25}+\frac{x^{2}}{3}=1

si graficamos la hipérbola, queda como en el documento adjunto.

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