Question

**GIVING MEDAL**
Please help, I’m very confused :(

What is the standard form of the equation of the conic given by
2x^2 + 2y^2 + 4x – 12y – 22 = 0

Answer 1

Choices:

(x + 1)^2/21 – (y - 3)^2/21 = 1

(x + 1)^2/21 + (y - 3)^2/21 = 1

(x - 3)^2/21 + (y + 1)^2/21 = 1

(x - 1)^2/7 + (y + 3)^2/3 = 1

Answer 2

combine x term and y term seperatly

Answer 3

first take 2 common

Answer 4

after taking 2 commonweare left with
x^2+y^2+2x-6y-11=0

Answer 5

x^2+2*x*1+1-1+y^2-2*3*y+9-9-11=0
x^2+2*x*1=(x+1)^2
y^2-2*3*y+9=(y-3)^2
(x+1)^2+(y-3)^2=11+9+1=21
(x+1)^2/21+(y-3)^2/21=1

Answer 6

Ah, thank you so much! I actually understand it now!