Identify the graph of the equation. Then find 0 to the nearest degree.
2x² + 4xy + 4y2 = 0

It is a type of conic section, but there is only one point .

This represent a single point (0, 0)  in the real plane.

This is a complex graph x = i*y  - y,   there are imaginary numbers.

Step-by-step explanation:

2 xx + 4xy +  4yy = 0

...xx + 2xy + y*y  + yy = 0

(x + y)^2  + yy = 0

if  x is a constant

then the cross sections would be parabolas in the zy-plane

... 2yy + 4xy  - 2xy  + xx = 0

(2yy + 4xy + 2xx -2xx) - 2xy + xx = 0

2(y + x)^2 - 2xx - 2xy + xx = 0

2(y + x)^2 - xx - 2xy  = 0

2(y + x)^2 - xx - 2xy  +yy  - yy = 0

...

(x + y)^2  + yy = 0

x = i*y  - y

if  x = r cos a

y = r sin a

cos a )^2  + cos a sina  + sin a)^2  +  sin a)^2 = 0

cos a sin a + sin a)^2 = -1

vbbv

step-by-step explanation:

the correct answers i think are b. and d.

give me brainliest if it is correct!