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(A) The given equation is $x^2 + 2\sqrt{3}xy - y^2 = 2a^2$. Since the axes are rotated through an angle $	heta = 30^{\circ}$,the transformation equat

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$X^2 - Y^2 = a^2$

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(A) The given equation is $x^2 + 2\sqrt{3}xy - y^2 = 2a^2$. Since the axes are rotated through an angle $	heta = 30^{\circ}$,the transformation equations are: $x = X \cos 30^{\circ} - Y \sin 30^{\circ} = \frac{\sqrt{3}X - Y}{2}$ $y = X \sin 30^{\circ} + Y \cos 30^{\circ} = \frac{X + \sqrt{3}Y}{2}$ Substituting these into the given equation: $\left(\frac{\sqrt{3}X - Y}{2}ight)^2 + 2\sqrt{3}\left(\frac{\sqrt{3}X - Y}{2}ight)\left(\frac{X + \sqrt{3}Y}{2}ight) - \left(\frac{X + \sqrt{3}Y}{2}ight)^2 = 2a^2$ Multiplying by $4$: $(\sqrt{3}X - Y)^2 + 2\sqrt{3}(\sqrt{3}X^2 + 3XY - XY - \sqrt{3}Y^2) - (X + \sqrt{3}Y)^2 = 8a^2$ $(3X^2 - 2\sqrt{3}XY + Y^2) + 2\sqrt{3}(\sqrt{3}X^2 + 2XY - \sqrt{3}Y^2) - (X^2 + 2\sqrt{3}XY + 3Y^2) = 8a^2$ $3X^2 - 2\sqrt{3}XY + Y^2 + 6X^2 + 4\sqrt{3}XY - 6Y^2 - X^2 - 2\sqrt{3}XY - 3Y^2 = 8a^2$ Combining like terms: $(3 + 6 - 1)X^2 + (-2\sqrt{3} + 4\sqrt{3} - 2\sqrt{3})XY + (1 - 6 - 3)Y^2 = 8a^2$ $8X^2 - 8Y^2 = 8a^2$ $X^2 - Y^2 = a^2$

Suppose the new axes $X, Y$ are generated by rotating the coordinate axes $x, y$ about the origin through an angle of $30^{\circ}$ in the anti-clockwise direction. Then,the transformed equation of $x^2+2 \sqrt{3} xy - y^2 = 2a^2$ with respect to the new axes $X, Y$ is

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