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3x^2-12y^2+6x+48y-93=0

Given the equation: 3x^2-12y^2+6x+48y-93=0 determine: 
a: The center C
b: the two vertices
c: the slopes of the asymptotes
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1 Answer

This is equation of hyperbola in general form which you need to change it to standard form to see the information asked. You need to complete the square for both x and y.
 
3(x^2+2x) - 12(y^2-4y) = 93
3(x^2+2x+1) - 12(y^2-4y+4) = 93 + 3 -48
3(x+1)^2 - 12(y-2)^2 = 48
 
divide both side by 48
 
(x+1)^2/16 - (y-2)^2/4 = 1
 
now center is at (-1, 2)
vertices (-5,2) and (3,2)
Slopes of the asymptotes - 1/2 and 1/2