Question

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

Answer 1

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 :)