Question 1121260
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I'm going to assume that you can get to the center, foci, vertices, and equations of the asymptotes if you have the Standard form.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  4x^2\ -\ 5y^2\ +\ 32x\ +\ 30y\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  4\(x^2\ +\ 8x)\ -\ 5(y^2\ -\ 6y)\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  4\(x^2\ +\ 8x\ +\ 16)\ -\ 5(y^2\ -\ 6y\ +\ 9)\ =\ 1\ +\ 64\ -\ 45]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4(x\ +\ 4)^2\ -\ 5(y\ -\ 3)^2\ =\ 20]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(x\ +\ 4)^2}{5\cdot 20}\ -\ \frac{(y\ -\ 3)^2}{4\cdot 20}\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(x\ +\ 4)^2}{10^2}\ -\ \frac{(y\ -\ 3)^2}{(4\sqrt{5})^2}\ =\ 1]
								
								
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 \ \ \ \ \ \ \ \ \ \  
								
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