Howndar do I write x^2-4y^2-4x-8y=36 in standard form?
Hyperbola (x-2)^2-4(y-1)^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
To write the equation x^2 - 4y^2 - 4x - 8y = 36 in standard form, you need to rearrange the equation so that the x and y terms are grouped separately and the constant term is on the other side of the equation. The standard form of an equation is Ax^2 + By^2 + Cx + Dy + E = 0, where A, B, C, D, and E are constants.
Here's a step-by-step process to convert the given equation into standard form:
1. Begin by moving the constant term (36) to the other side of the equation:
x^2 - 4y^2 - 4x - 8y - 36 = 0
2. Next, rearrange the x and y terms by grouping them separately:
(x^2 - 4x) - (4y^2 + 8y) - 36 = 0
3. Complete the square for the x and y terms individually:
For the x terms:
- Take half of the coefficient of x, which is -4, and square it: (-4/2)^2 = 4
- Add this value to both sides of the equation: (x^2 - 4x + 4) - (4y^2 + 8y) - 36 + 4 = 0
- Rewrite the x terms as a perfect square: (x - 2)^2 - (4y^2 + 8y) - 32 = 0
For the y terms:
- Take half of the coefficient of y, which is -8, and square it: (-8/2)^2 = 16
- Add this value to both sides of the equation: (x - 2)^2 - (4y^2 + 8y + 16) - 32 + 16 = 0
- Rewrite the y terms as a perfect square: (x - 2)^2 - 4(y + 2)^2 - 16 = 0
4. Finally, simplify and rewrite the equation in standard form:
(x - 2)^2 - 4(y + 2)^2 - 48 = 0
The equation x^2 - 4y^2 - 4x - 8y = 36 can be written in standard form as (x - 2)^2 - 4(y + 2)^2 - 48 = 0.