SOLUTION: hallar la ecuacion de las hiperbolas determinadas por: 1) vertices (+-1,0) asíntotas y=+-5x 2) focos (0,+-6) pasa por P=(-5,9) 3) focos (0,+-1) longitud eje real:1 4) asíntot

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Question 1133039: hallar la ecuacion de las hiperbolas determinadas por:
1) vertices (+-1,0) asíntotas y=+-5x
2) focos (0,+-6) pasa por P=(-5,9)
3) focos (0,+-1) longitud eje real:1
4) asíntotas y= +- x/2 pasa por el punto de coordenadas (5,2)

Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!

Standard form:
->Transverse axis is horizontal
-.>Transverse axis is vertical

1)
given:
Vertices: (,), (,) =>Semi-major axis length:


First asymptote:
Second asymptote:
=> Transverse axis is vertical
Standard form:
intersection of the asymptotes is at origin:
so, center is at: (,) => and
=>
From the original equations of asymptotes, you can determine the slopes of the asymptotes to be and
since and , we have

=>









2.


2) focos (0,+-6) pasa por P=(-5,9)

the coordinates of the foci are (,±)
foci lie on y axis, so your formula is:

=> or



P=(,)
Solved by pluggable solver: Distance Between 2 points
The distance formula is . Plug in the numbers,

The distance is 15.8113883008419.



=> round it to
Solved by pluggable solver: Distance Between 2 points
The distance formula is . Plug in the numbers,

The distance is 5.8309518948453.



=> round it to








3) focos (0,+-1) longitud eje real:1
longitud eje real:1=> the length transverse axis is
=>
=>
the coordinates of the foci are (,±) => =± and
foci lie on axis, so your formula is

=> or







4) asíntotas = ± pasa por el punto de coordenadas (5,2)


= ±
intersection of the asymptotes is at origin:
so, center is at: (,) => and
asymptotesup-down = ±


since passes through (,)












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