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August 23, 2014

August 23, 2014

Posted by **saad** on Wednesday, August 17, 2011 at 1:13am.

12x^2-18y^2-18x-12y+12=0

- solving conics -
**Reiny**, Wednesday, August 17, 2011 at 8:19amlet's complete the square to get it into standard form

12(x^2 - (3/2)x + .....) - 18(y^2 + (2/3)y + ...) = -12

12(x^2 - (3/2)x +**9/16**) - 18(y^2 + (2/3)y +**1/9**) = -12 +**9/16**+**1/9**

12(x - 3/4)^2 - 18(y + 1/3)^2 = -1631/144

divide by 1631/144

(x-3/4)^2/(1728/1631) - (y+1/3)^2/(2592/1631) = -1

I am sure you can determine the properties of the vertical hyperbola from there.

Check my arithmetic.

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