Posted by Tori on Monday, December 3, 2012 at 2:36am.
I am supposed to solve these conics, can you please help :)I am supposed to name the center and the vertices.
1. 3y^22x^2+12x+24+24y=0
2.16x^24y^2=48x20y+57

PreCalculus  Damon, Monday, December 3, 2012 at 3:42am
complete square again and again
3 y^2 + 24 y = 2 x^2 12 x  24
divide by 3
y^2 + 8 y = (2/3) x^2  4 x  8
add square of half of 8
y^2 + 8 y + 16 = (2/3) x^2  4 x + 8
or
(y+4)^2 = 
now multiply by (3/2)
(3/2)(y+4)^2 = x^2  6 x + 12
move 12 over
(3/2)(y+4)^2 12 = x^2  6 x
add square of half of 6
(3/2)(y+4)^2 3 = x^2  6 x + 9
or
(3/2)(y+4)^2 3 = (x3)^2
or
(3/2)(y+4)^2  (x3)^2 = 3
(y+4)^2 /2  (x3)^2/3 = 1
Patience, patience :)
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