Posted by Alfred on Monday, December 5, 2011 at 10:06pm.
complete the square
x^2 + 6x + 9 - 3y^2 = 9
(x+3)^2 - 3y^2 = 9
divide by 9
(x+3)^2 /9 - y^2/3 = 1
centre (-3,0)
a = 3, b = √3
vertices at (3,0) and (-3,0)
faintly sketch a rectangle with corners
at (±3,±√3)
draw in the diagonals and extend them
sketch the hyperbola with the diagonals as asymptotes and the vertices as noted above
except the whole thing is shifted 3 units to the left - because it's (x+3)^2
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