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

Question

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

Answer 1

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