Asked by JessieLynne
Could someone please show me how to write this equation in standard form?
x^2 + y^2 - 6x+8y+9=0
x^2 + y^2 - 6x+8y+9=0
Answers
Answered by
Steve
That is the standard general form. Now, if you want it as a conic section in standard form, group the x and y terms:
(x^2 - 6x) + (y^2 + 8y) + 9 = 0
Now, make each parenthesized group a perfect square, then subtract off whatever you had to add in:
(x^2 - 6x + 9) + (y^2 + 8y + 16) + 9-9-16 = 0
(x-3)^2 + (y-4)^2 = 16
(x^2 - 6x) + (y^2 + 8y) + 9 = 0
Now, make each parenthesized group a perfect square, then subtract off whatever you had to add in:
(x^2 - 6x + 9) + (y^2 + 8y + 16) + 9-9-16 = 0
(x-3)^2 + (y-4)^2 = 16
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