Suppose the new axes X,Y are generated by rotating the coordinate axes x,y about the origin through an angle of $30^{0}$ in the anticlockwise
direction. Then the transformed equation of $x^{2} + 2 \sqrt{3} x y - y^{2} = 2 a^{2}$ with respect to new axes X,Y is

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

$X^{2} + 2 \sqrt{3} X Y - Y^{2} = 2 a^{2}$

$X^{2} + Y^{2} = 2 a^{2}$

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

answer is A.

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Detailed Solution

$Given θ = 30^{0} Now x = X \cos 30^{0} - Y \sin 30^{0} = \frac{\sqrt{3} X}{2} - \frac{Y}{2} = \frac{\sqrt{3} X - Y}{2} and y = Xsin 30^{0} + Y \cos 30^{0} = \frac{X}{2} + \frac{\sqrt{3} Y}{2} = \frac{X + \sqrt{3} Y}{2}$

$Now transformed equation of x^{2} + 2 \sqrt{3} xy - y^{2} = 2 a^{2} {(\frac{\sqrt{3} X - Y}{2})}^{2} + 2 \sqrt{3} (\frac{\sqrt{3} X - Y}{2}) (\frac{X + Y}{2}) - {(\frac{X + \sqrt{3} Y}{2})}^{2} = 2 a^{2} \Rightarrow {\frac{(\sqrt{3} X - Y)}{4}}^{2} + 2 \sqrt{3} \frac{[\sqrt{3} X^{2} + 2 X Y - \sqrt{3} Y^{2}]}{4} - \frac{{(X + \sqrt{3} Y)}^{2}}{4} = 2 a^{2} \Rightarrow 3 X^{2} + Y^{2} - 2 \sqrt{3} X Y + 6 X^{2} + 4 \sqrt{3} X Y - 6 Y^{2} - X^{2} - 3 Y^{2} - 2 \sqrt{3} X Y = 8 a^{2} \Rightarrow 8 X^{2} - 8 Y^{2} = 8 a^{2} \Rightarrow X^{2} - Y^{2} = a^{2}$