SOLUTION: For the hyperbola, determine the a) coordinates of the center b) directions and lengths of both axes c) coordinates of the vertices d) slopes of the asymptotes https://i.ibb.c

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: For the hyperbola, determine the a) coordinates of the center b) directions and lengths of both axes c) coordinates of the vertices d) slopes of the asymptotes https://i.ibb.c      Log On


   



Question 1204680: For the hyperbola, determine the
a) coordinates of the center
b) directions and lengths of both axes
c) coordinates of the vertices
d) slopes of the asymptotes
https://i.ibb.co/6YZnZHT/1.jpg
I don’t know why this graph is confusing me. It could be the seemingly asymmetric look of the asymptotes. I don’t know how to trace a box in the center with asymptotes that appear slanted and the two parts of the hyperbola that appear different to me.

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

The standard form of the equation of a hyperbola with center (0,0) and transverse axis on the x-axis is
x%5E2%2Fa%5E2-y%5E2%2Fb%5E2=1
we know that the coordinates of the center are
(h,k)=(0,0)
and the vertices are
(-a,0)=(-2,0) and (a,0)=(2,0)
=> a=2
since the transverse axis is horizontal, asymptote formula will be:

y%28b%2Fa%29x
from graph we see that one asymptote passes through the origin (0,0) and (2,-1)
y=mx
slope is:
m=%28-1-0%29%2F%282-0%29=-1%2F2
equation is:
y=-%281%2F2%29x
the other asymptote passes through origin and point (2,1)
slope is:
m=%281-0%29%2F%282-0%29=1%2F2
equation is:
y=%281%2F2%29x
then, since y%28b%2Fa%29x, b=1
and your equation of hyperbola is:
x%5E2%2F2%5E2-y%5E2%2F1%5E2=1
x%5E2%2F4-y%5E2%2F1=1
x%5E2%2F4-y%5E2=1

For the hyperbola, determine the
a) coordinates of the center
(0,0)
b) directions and lengths of both axes
the transverse axis is horizontal

the length of the axes is 2a+=2%2A2=4 and the length of conjugate axis is 2b=2%2A1=2

c) coordinates of the vertices
(-2,0) and (2,0)

d) slopes of the asymptotes

from graph we see that one asymptote passes through the origin (0,0) and (2,-1)
y=mx
slope is:
m=%28-1-0%29%2F%282-0%29=-1%2F2
equation is:
y=-%281%2F2%29x
the other asymptote passes through origin and point (2,1)
slope is:
m=%281-0%29%2F%282-0%29=1%2F2
equation is:
y=%281%2F2%29x