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?

http://i152.photobucket.com/albums/s...Hyperbola1.jpg

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- February 16th 2007, 07:22 PM^_^Engineer_Adam^_^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?

http://i152.photobucket.com/albums/s...Hyperbola1.jpg - February 16th 2007, 10:43 PMCaptainBlack
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 - February 17th 2007, 02:46 AMticbol
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.