# Algebra II

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Howndar do I write x^2-4y^2-4x-8y=36 in standard form?

• Algebra II -

Hyperbola (x-2)^2-4(y-1)^2=36

I have that one on my exam too.

• Algebra II -

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.

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.

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( 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

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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