SOLUTION: 3x2+45x-12y2+60y=-60

Question 29081: 3x2+45x-12y2+60y=-60
Answer by sdmmadam@yahoo.com(530) (Show Source): You can put this solution on YOUR website!
3x^2+45x-12y^2+60y=-60
x^2+15x-4y^2+20y=-20 (dividing through out by 3)
[x^2+2X(15/2)x +(15/2)^2-(15/2)^2]-4(y^2-5y)=-20
(x+15/2)^2 - 225/4-4[y^2-2X(5/2)y+(5/2)^2-(5/2)^2]= -20
(x+15/2)^2 -4[(y-5/2)^2-(5/2)^2]= -20+ 225/4
(x+15/2)^2 -4(y-5/2)^2+4X(25/4)= (-80+225)/4
(x+15/2)^2 -4(y-5/2)^2+25= (-80+225)/4
(x+15/2)^2 -4(y-5/2)^2 = (-80+225)/4-25
(x+15/2)^2 -4(y-5/2)^2 = (145-100)/4
(x+15/2)^2 -4(y-5/2)^2 = 45/4
Dividing by 45/4 that is multiplying by (4/45)
{[x-(-15/2)]^2}/(45/4) -4{[y-(5/2)^2}/(45/4) =1
{[x-(-15/2)]^2}/(45/4) -{[y-(5/2)^2}/(45/16) =1
which is of the form (x-h)^2/(a^2) -(y-k)^2/(b^2) = 1 which is the equation to a hyperbola in the general form with centre=(h,k) = (-15/2,5/2) (a point in the second quadrant) with axes respectively parallel to the rectangular coordinate axes of reference,
the transverse axis parallel to the x axis given by y = (5/2)
the conjugate axis parallel to the y axis given by x = (-15/2)
a= semi transverse length =sqrt(45/4) = (3/2)sqrt(5)
b= semi conjugate length =sqrt(45/16) = (3/4)sqrt(5)
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