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October 31, 2014

October 31, 2014

Posted by **Blake** on Monday, April 2, 2007 at 1:29pm.

12x^2+y^2+6x-9=0

not sure what you mean by "solve this conic"

You want to know what kind of conic? ellipse

to find out its properties you have to change it into standard form (x-h)^2/a^2 + (y-k)^2/b^2 = 1

I would first divide by 12

x^2 + y^2/12 + x/2 = 9/12

complete the square for the x terms, adding the same quantity to both sides

x^2 + x/2 + 1/16 + y^2/12 = 3/4 + 1/16

(X + 1/4)^2 + Y^2/12 = 13/16

Divide each term by 13/16

(x + 1/4)^2/(13/16) + y^2/(26/16) = 1

so we have an ellipse with centre at (-1/4,0), with a = √13 /4 and b = √26 / 4

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