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Identify the graph of 2x^2 + 4xy + 4y^2 = 0 then find theta to the nearest degree.

User Shammara
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2 Answers

4 votes

Answer:

Ellipse; -32deg

Explanation:

Got it right on Edge

User Nhatnq
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7 votes

Answer: 2x2 + 4xy + 4y2 = 0

2 xx + 4xy + 4yy = 0

...xx + 2xy + y*y + yy = 0

(x + y)^2 + yy = 0

if x is a constant

... 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

Explanation:

User Anuja P
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4.3k points