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

Algebra ->  Quadratic-relations-and-conic-sections -> 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      Log On


   

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(20849) About Me  (Show Source):
You can put this solution on YOUR website!

Standard form:
%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1->Transverse axis is horizontal
%28y-h%29%5E2%2Fa%5E2-%28x-k%29%5E2%2Fb%5E2=1-.>Transverse axis is vertical

1)
given:
Vertices: (0,-6), (0,6) =>Semi-major axis length: a=6
%28x-h%29%5E2%2F6%5E2-%28y-k%29%5E2%2Fb%5E2=1
%28x-h%29%5E2%2F36-%28y-k%29%5E2%2Fb%5E2=1
First asymptote: y=-5x
Second asymptote: y=5x
=> Transverse axis is vertical
Standard form: %28y-h%29%5E2%2Fa%5E2-%28x-k%29%5E2%2Fb%5E2=1
intersection of the asymptotes is at origin:
so, center is at: (0,0) => h=0+and k=0
=>y%5E2%2Fa%5E2-x%5E2%2Fb%5E2=1
From the original equations of asymptotes, you can determine the slopes of the asymptotes to be m=5 and m=-5
since m=a%2Fb and a=6, we have
5=6%2Fb
=>b=6%2F5
x%5E2%2F36-y%5E2%2F%286%2F5%29%5E2=1
x%5E2%2F36-y%5E2%2F%2836%2F25%29=1
x%5E2%2F36-25y%5E2%2F36=1






2.


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

the coordinates of the foci are (0,±c)
foci lie on y axis, so your formula is:
y%5E2%2Fa%5E2-x%5E2%2Fb%5E2=1
=> c=6 or c=-6
6%5E2=a%5E2%2Bb%5E2
36=a%5E2%2Bb%5E2
b%5E2=36-a%5E2
P=(-5,9)
Solved by pluggable solver: Distance Between 2 points
The distance formula is sqrt%28%28%28x%5B2%5D-x%5B1%5D%29%5E2%29%2B%28%28y%5B2%5D-y%5B1%5D%29%5E2%29%29. Plug in the numbers,
sqrt%28%28%28-5-%280%29%29%5E2%29%2B%28%289-%28-6%29%29%5E2%29%29
sqrt%28-5%5E2%2B15%5E2%29 The distance is 15.8113883008419.



d%5B1%5D=15.8113883008419=> round it to d%5B1%5D=16
Solved by pluggable solver: Distance Between 2 points
The distance formula is sqrt%28%28%28x%5B2%5D-x%5B1%5D%29%5E2%29%2B%28%28y%5B2%5D-y%5B1%5D%29%5E2%29%29. Plug in the numbers,
sqrt%28%28%28-5-%280%29%29%5E2%29%2B%28%289-%286%29%29%5E2%29%29
sqrt%28-5%5E2%2B3%5E2%29 The distance is 5.8309518948453.



d%5B2%5D=5.8309518948453=> round it to d%5B2%5D=6
d%5B1%5D-d%5B2%5D=2a
2a=16-6=10
a=10%2F2=5
b%5E2=36-5%5E2
b%5E2=36-25
b%5E2+=+11
y%5E2%2F25-x%5E2%2F11=1

3) focos (0,+-1) longitud eje real:1
longitud eje real:1=> the length transverse axis is 1
=>2a=1
=>a=1%2F2
the coordinates of the foci are (0,±c) => c=±1 and
foci lie on y axis, so your formula is
y%5E2%2Fa%5E2-x%5E2%2Fb%5E2=1
=> c=1 or c=-1
1%5E2=%281%2F2%29%5E2%2Bb%5E2
1=1%2F4%2Bb%5E2
b%5E2=1-1%2F4
b%5E2=3%2F4
y%5E2%2F%281%2F4%29-x%5E2%2F%283%2F4%29=1
4y%5E2-4x%5E2%2F3=1

4) asíntotas y= ± x%2F2 pasa por el punto de coordenadas (5,2)


y= ± %281%2F2%29x
intersection of the asymptotes is at origin:
so, center is at: (0,0) => h=0+and k=0
asymptotesup-down y= ± %28a%2Fb%29x
a%2Fb=1%2F2
2a=b
since passes through (5,2)
5%5E2%2Fa%5E2-2%5E2%2Fb%5E2=1
25%2Fa%5E2-4%2F%282a%29%5E2=1
25%2Fa%5E2-4%2F4a%5E2=1
25%2Fa%5E2-1%2Fa%5E2=1
24%2Fa%5E2=1
a%5E2=24
b=2a=2sqrt%2824%29
b%5E2=4%2A24
b%5E2=96

y%5E2%2F24-x%5E2%2F96=1