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(A) Since the coordinate axes are rotated through an angle $	heta = 45^{\circ}$,we replace $(x, y)$ by $(x \cos 45^{\circ} - y \sin 45^{\circ}, x \si

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

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(A) Since the coordinate axes are rotated through an angle $	heta = 45^{\circ}$,we replace $(x, y)$ by $(x \cos 45^{\circ} - y \sin 45^{\circ}, x \sin 45^{\circ} + y \cos 45^{\circ})$.This simplifies to $\left(\frac{x-y}{\sqrt{2}}, \frac{x+y}{\sqrt{2}}ight)$.Substituting these into the equation $3x^2 + 3y^2 + 2xy - 2 = 0$:$3\left(\frac{x-y}{\sqrt{2}}ight)^2 + 3\left(\frac{x+y}{\sqrt{2}}ight)^2 + 2\left(\frac{x-y}{\sqrt{2}}ight)\left(\frac{x+y}{\sqrt{2}}ight) - 2 = 0$$\Rightarrow \frac{3}{2}(x^2 + y^2 - 2xy) + \frac{3}{2}(x^2 + y^2 + 2xy) + (x^2 - y^2) - 2 = 0$$\Rightarrow \frac{3}{2}(2x^2 + 2y^2) + x^2 - y^2 - 2 = 0$$\Rightarrow 3x^2 + 3y^2 + x^2 - y^2 - 2 = 0$$\Rightarrow 4x^2 + 2y^2 = 2$$\Rightarrow 2x^2 + y^2 = 1$.

If the coordinate axes are rotated in the positive direction by $45^{\circ}$ without changing the origin,then the transformed equation of $3x^2 + 3y^2 + 2xy - 2 = 0$ is

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