# Thread: Need help on this hyperbola problem:

1. ## Need help on this hyperbola problem:

FInd the equation that satisfy the given conditions:
See image:
(have only 1 given which is (1,1))
Does it mean that the vertices and the Foci are the same?

2. Originally Posted by ^_^Engineer_Adam^_^
FInd the equation that satisfy the given conditions:
See image:
(have only 1 given which is (1,1))
Does it mean that the vertices and the Foci are the same?
The condition and diagram mean that the hyperbola passes through the
points (0,1), and (0,-1), where the curve is horizontal.

Given the orientation of the hyperbola, its equation is of the form:

-a x^2 + b y^2 = 1

for some a, b>0. Then the condition that it passes through (0,1) gives:

b=1,

and as the asymptotes are y=+/- x, we also have a=b, so the hyperbola
is:

x^2 - y^2 = 1.

RonL

3. Originally Posted by ^_^Engineer_Adam^_^
FInd the equation that satisfy the given conditions:
See image:
(have only 1 given which is (1,1))
Does it mean that the vertices and the Foci are the same?
The hyperbola opens up and down, meaning the transverse axis is vertical. So the standard form of the equation of that hyperbola is:
[(y-k)^2 /(a^2)] -[(x-h)^2 /(b^2] = 1 ----------(i)
where
(h,k) is the center of the hyperbola
a is the semi-major axis or semi-transverse axis -----it is on the focal axis, so "a" is vertical here, or is parallel to the y-axis.
b is semi-minor axis or semi-conjugate axis ----parallel to the x-axis.

The center is shown as at (0,0), so Eq.(i) becomes
[y^2 / a^2] -[x^2 / b^2] = 1 -------------------(ii)

The given point (1,1) means (b,a) here. Or b = 1, and a = 1.
Hence, (ii) becomes
[y^2 / 1^2] -[x^2 / 1^2] = 1
y^2 -x^2 = 1 -----------------the hyperbola, answer.