Question

What are the asymptotes of the hyperbola given by the equation
y^2/1- x^2/121 = 1?

Answer 1

You can use the same formula as the last question since it's another vertical hyperbola. This time h and k are zero, a^2=1 and b^2=121.

Answer 2

\[y=\pm \frac{ a }{ b }(x-h)+k \]

Answer 3

hyperbola with vertical transverse axis. The standard form is:
\[\frac{ (y-k)^2 }{ a^2 } - \frac{ (x-h)^2 }{ b^2 } = 1\]
where (h,k) = (x,y) respectively. And are the coordinates of center.
For your problem:
center: (0,0)
a = 1
b = 121 = 11
(it's 11 because it's b^2, hence, b = 11. You can't have b^2).
Now, remember, asymptotes are straight lines that go thru the center (0,0). While your standard form of equation for a straight line is defined by:
y = mx + b
m = slope
b = y-intercept
Slopes of asymptotes are:
\[\pm \frac{ a }{ b } = \frac{ 1 }{ 11 }\]
hence:
\[y = \pm \frac{ x }{ 11 } + b\]
since your center is at (0,0), y-intercept = 0
b = 0
and your equation of asymptotes is:
\[y = \pm \frac{ x }{ 11 }\]

Answer 4

choices are:

y=121x

y=11x

y= 1/11x

y= 1/121x

Answer 5

@abb0t which one is it?

Answer 6

which one does
\[y = \pm \frac{ x }{ 11 } \] look like?

Answer 7

\[y = \pm \frac{ x }{ 11 } = \pm \frac{ 1 }{ 11 }x\]