Question 1120388
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There is an *[tex \Large x^2] term and a *[tex \Large y^2] term.  Eliminate Parabola.


The coefficients on *[tex \Large x^2] and *[tex \Large y^2] are different.  Eliminate a circle.


The signs on the *[tex \Large x^2] and *[tex \Large y^2] terms are opposite.  Eliminate an ellipse.


Therefore, it must be a hyperbola.


Complete the square:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2\  -\ 2y^2\ +\ 2x\ +\ 8y\ -\ 7\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2\  -\ 2y^2\ +\ 2x\ +\ 8y\ =\ 7]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2\ +\ 2x\ -\ 2y^2\ +\ 8y\ =\ 7]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2\ +\ 2x\ +\ 1\ -\ 2(y^2\ -\ 4y\ +\ 4)=\ 7\ +\ 1\ -\ 8]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (x\ +\ 1)^2\ -\ 2(y\ -\ 2)^2\ =\ 0]


No conic exists.  The graph is two straight lines.
								
								
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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