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^2-2x^2+12x+24+24y=0

2.-16x^2-4y^2=48x-20y+57

- Pre-Calculus -
**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 = (x-3)^2

or

(3/2)(y+4)^2 - (x-3)^2 = 3

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

Patience, patience :)

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