The transformed equation of $3 x^{2} + 3 y^{2} + 2 x y = 2$ when the coordinate axes are rotated through an angle of $45^{o}$ is

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$x^{2} + 2 y^{2} = 1$

$x^{2} + y^{2} = 1$

$x^{2} + 3 y^{2} = 1$

$2 x^{2} + y^{2} = 1$

answer is B.

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Detailed Solution

$Given angle of rotation θ = 45^{0} Given original equation 3 x^{2} + 3 y^{2} + 2 xy - 2 = 0 Now x = X \cos 45^{0} - Y \sin 45^{0} = X (\frac{1}{\sqrt{2}}) - Y (\frac{1}{\sqrt{2}}) = \frac{X - Y}{\sqrt{2}} y = X \sin 45^{0} + Y \cos 45^{0} = X (\frac{1}{\sqrt{2}}) + Y (\frac{1}{\sqrt{2}}) y = \frac{X + Y}{\sqrt{2}}$

$Now transformed equation is 3 {(\frac{X - Y}{\sqrt{2}})}^{2} + 3 {(\frac{X + Y}{\sqrt{2}})}^{2} + 2 (\frac{X - Y}{\sqrt{2}}) (\frac{X + Y}{\sqrt{2}}) - 2 = 0 \Rightarrow \frac{3 {(X - Y)}^{2}}{2} + \frac{3 {(X + Y)}^{2}}{2} + 2 (\frac{X^{2} - Y^{2}}{2}) - 2 = 0 \Rightarrow \frac{3 (X^{2} + Y^{2} - 2 X Y) + 3 (X^{2} + Y^{2} + 2 X Y) + 2 (X^{2} - Y^{2}) - 4}{2} = 0 \Rightarrow 3 X^{2} + 3 Y^{2} - 6 X Y + 3 X^{2} + 3 Y^{2} + 6 X Y + 2 X^{2} - 2 Y^{2} - 4 = 0 \Rightarrow 8 X^{2} + 4 Y^{2} = 4 \Rightarrow 2 X^{2} + Y^{2} = 1$