Algebra II
posted by Sharon .
Howndar do I write x^24y^24x8y=36 in standard form?

Hyperbola (x2)^24(y1)^2=36
I have that one on my exam too. 
The General Equation for a Conic Sections:
A x ^ 2 + B x y + C y ^ 2 + D x + E y + F = 0
In this case:
x ^ 2  4 y ^ 2  4 x  8 y = 36 Subtract 36 to both sides
x ^ 2  4 y ^ 2  4 x  8 y  36 = 0
1 * x ^ 2 + 0 * B x y  4 * y ^ 2  4 * x  8 * y  36 = 0
A = 1
B = 0
C =  4
D =  4
E =  8
F =  36
The discriminant B ^ 2  4 A C will identify which conic section it is.
If the discriminant is positive, the section is a hyperbola.
If it is negative, the section is an ellipse.
If it is zero, the section is a parabola.
B ^ 2  4 A C = 0 ^ 2  4 * 1 * (  4 ) = 0 + 16 = 16
The discriminant is positive.
Your line is a hyperbola.
Equation of hyperbola in standard form :
( x  h ) ^ 2 / a ^ 2  ( y  k ) ^ 2 / b ^ 2 = 1
x ^ 2  4 y ^ 2  4 x  8 y = 36
x ^ 2  4 x  4 y ^ 2  8 y = 36
( x ^ 2  4 x )  4 ( y ^ 2 + 2 y ) = 36
The process involves completing the square separately for the x and y variables.
____________________________________
( x  2 ) ^ 2 = x ^ 2  2 * x * 2 + 2 ^ 2
( x  2 ) ^ 2 = x ^ 2  4 x + 4 Subtract 4 to both sides
( x  2 ) ^ 2  4 = x ^ 2  4 x + 4  4
( x  2 ) ^ 2  4 = x ^ 2  4 x
x ^ 2  4 x = ( x  2 ) ^ 2  4
( y + 1 ) ^ 2 = y ^ 2 + 2 * y * 1 + 1 ^ 2
( y + 1 ) ^ 2 = y ^ 2 + 2 y + 1 Subtract 1 to both sides
( y + 1 ) ^ 2  1 = y ^ 2 + 2 y + 1  1
( y + 1 ) ^ 2  1 = y ^ 2 + 2 y
y ^ 2 + 2 y = ( y + 1 ) ^ 2  1
 4 * ( y ^ 2 + 2 y ) =  4 * [ ( y + 1 ) ^ 2  1 ]
 4 * ( y ^ 2 + 2 y ) =  4 ( y + 1 ) ^ 2 + 4
__________________________
x ^ 2  4 x  4 y ^ 2  8 y = 36
x ^ 2  4 x  4 ( y ^ 2 + 2 y ) = 36
( x ^ 2  4 x )  4 ( y ^ 2 + 2 y ) = 36
( x  2 ) ^ 2  4  4 ( y + 1 ) ^ 2 + 4 = 36
( x  2 ) ^ 2  4 ( y + 1 ) ^ 2 = 36 Divide both sides by 36
( x  2 ) ^ 2 / 36  4 ( y + 1 ) ^ 2 / 36 = 1
( x  2 ) ^ 2 / 36  4 ( y + 1 ) ^ 2 / ( 4 * 9 ) = 1
( x  2 ) ^ 2 / 36  ( y + 1 ) ^ 2 / 9 = 1
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