Question 1163046
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The length of a latus rectum of hyperbola *[tex \Large \frac{x^2}{a^2}\ -\ \frac{y^2}{b^2}\ =\ 1] is *[tex \Large \frac{2b^2}{a}]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
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