# CONVERTING FROM GENERAL FORM TO STANDARD FORM FOR PARABOLA

Ax2 + Bx + Cy + D = 0

Ay2 + By + Cx + D = 0

Observing the equations above, the general equation of parabola will have square for any one of the variable either x or y.

To convert this equation to standard form, we have to remember

y = a(x - h)2 + k

x = a(y - k)2 + h

Write the following equations in standard form :

Problem 1 :

y = 2x2 - 12 x + 6

Solution :

y = 2x2 - 12 x + 6

Factoring 2 from x terms,

y = 2(x2 - 6 x) + 6

y = 2[x2 - 2 x (3) + 32 - 32] + 6

y = 2[(x - 3)2 - 9] + 6

Distributing 2, we get

y = 2(x - 3)2 - 18 + 6

y = 2(x - 3)2 - 12

Problem 2 :

y = x2 - 6 x + 11

Solution :

y = x2 - 6 x + 11

y = [x2 - 2 x (3) + 32 - 32] + 11

y = [(x - 3)2 - 32] + 11

y = [(x - 3)2 - 9] + 11

y = (x - 3)2 - 9 + 11

y = (x - 3)2 + 2

Problem 3 :

x = y2 + 14y + 20

Solution :

x = y2 + 14y + 20

x = [y2 + 2 x (7) + 72 - 72] + 20

x = [(y + 7)2 - 72] + 20

x = (y + 7)2 - 49 + 20

x = (y + 7)2 - 29

Problem 4 :

y = (1/2) x2 + 12x - 8

Solution :

y = (1/2) x2 + 12x - 8

Factoring 1/2 from x2 and x, we get

y = (1/2)[x2 + 24x] - 8

y = (1/2)[x2 + 2 x (12) + 122 - 122] - 8

y = (1/2)[(x + 12)2  - 144] - 8

Distributing 1/2, we get

y = (1/2)(x + 12)2  - 72 - 8

y = (1/2)(x + 12)2  - 80

Problem 5 :

x = 3y2 + 5y - 9

Solution :

x = 3y2 + 5y - 9

Factoring 3 from y terms it is not possible, because the coefficient of y is not a multiple of 3.

x = 3[y2 + (5/3)y] - 9

Writing the coefficient of x as multiple of 2, we get

x = 3[y2 + 2(5/6)y + (5/6)2 - (5/6)2] - 9

x = 3[(y + (5/6))2 - (25/36)] - 9

x = 3(y + (5/6))2 - (25/12) - 9

x = 3(y + (5/6))2 + (-25 - 108)/12

x = 3(y + (5/6))2 + (-133)/12

Problem 6 :

Katie is finding the standard form of the equation

y = x2 + 6x + 4

Solution :

y = x2 + 6x + 4

y = x2 + 2 x (3) + 32 - 32 + 4

y = (x + 3)2 - 32 + 4

y = (x + 3)2 - 9 + 4

y = (x + 3)2 - 5

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