Question 194672
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First, put it in standard form.  Either:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1]



If you add -1 to both sides you have something equal to -1 so you have to multiply the whole thing by -1, that gives you:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -4x^2 + y^2 = 1 \ \ \Rightarrow\ \ y^2 - 4x^2 = 1 \ \ \Rightarrow\ \ \frac{(y-0)^2}{1^2}-\frac{(x-0)^2}{\left({1 \over 2}\right)^2}=1]


This hyperbola is centered at the origin and, because the y-term is the positive term, it opens up and down and the foci are on the y-axis at (0,c) where


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c = \pm\sqrt{a^2 + b^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c = \pm\sqrt{1 + {1 \over 4}} =\pm \frac{\sqrt{5}}{2}]


Hence, the foci are at *[tex \Large \left(0,\frac{\sqrt{5}}{2}\right)] and *[tex \Large \left(0,-\frac{\sqrt{5}}{2}\right)]





John
*[tex \LARGE e^{i\pi} + 1 = 0]
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