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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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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Unit 3 homework 6 Gina Wilson
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Answer:

5) The equation of the straight line is   2 x - y + 1 =0

6) The equation of the straight line is   x + y -5 =0

7) The equation of the straight line is   5 x + 6 y - 24 =0

8) The equation of the straight line is  x - 4 y -4 =0

9) The equation of the parallel line is 3x + y -19 =0

Step-by-step explanation:

5)

The equation of the straight line is

                         y - y_{1}  = m ( x - x_{1} )

          Slope of the line

                      m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

        Given points are (1,3) , ( -3,-5)

         m = \frac{-5-3  }{-3-1 } = \frac{-8}{-4} = 2

The equation of the straight line is

                         y - y_{1}  = m ( x - x_{1} )

                        y - 3 = 2 ( x - 1 )

                        y = 2x - 2 +3

                       2 x - y + 1 =0

The equation of the straight line is   2 x - y + 1 =0

  6)

The equation of the straight line is

                         y - y_{1}  = m ( x - x_{1} )

          Slope of the line

                      m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

        Given points are (1,4) , ( 6,-1)

         m = \frac{-1-(4)  }{6-1 } = \frac{-5}{5} = -1

The equation of the straight line is  

                         y - y_{1}  = m ( x - x_{1} )

                        y - 1 = -1 ( x - 4 )

                        y - 1 = - x +4

The equation of the straight line is   x + y -5 =0

7)

The equation of the straight line is

                         y - y_{1}  = m ( x - x_{1} )

          Slope of the line

                      m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

 Given points are (-12 , 14) , ( 6,-1)

         m = \frac{-1-(14)  }{6+12 } = \frac{-15}{18} = \frac{-5}{6}

         y - 14  = \frac{-5}{6}  ( x - (-12) )

       6( y - 14 ) = - 5 ( x +12 )

      6 y - 84 = - 5x -60

       5 x + 6 y  -84 + 60 =0

      5 x + 6 y - 24 =0

8)

The equation of the straight line is

                         y - y_{1}  = m ( x - x_{1} )

          Slope of the line

                      m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

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           y - (-2)  =\frac{1}{4} ( x - (-4) )

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                x - 4 y -4 =0

  9)

The equation of the line y = 3x + 6  is parallel to the line

3x + y + k =0 is passes through the point ( 4,7 )

⇒   3x + y + k =0

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⇒    k = -19

The equation of the parallel line is 3x + y -19 =0

     

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Step-by-step explanation:

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And then, find the distance.

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