Suppose the axes X and Y are obtained by rota | vedclass

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(D) The transformation equations for rotation of axes by an angle $	heta$ are $x = X \cos 	heta - Y \sin 	heta$ and $y = X \sin 	heta + Y \cos 	h

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(D) The transformation equations for rotation of axes by an angle $	heta$ are $x = X \cos 	heta - Y \sin 	heta$ and $y = X \sin 	heta + Y \cos 	heta$. Substituting these into the given equation $x^2 + 2\sqrt{3}xy - y^2 = 4a^2$: $(X \cos 	heta - Y \sin 	heta)^2 + 2\sqrt{3}(X \cos 	heta - Y \sin 	heta)(X \sin 	heta + Y \cos 	heta) - (X \sin 	heta + Y \cos 	heta)^2 = 4a^2$. Expanding the terms,the coefficient of the $XY$ term must be zero for the equation to take the form $X^2 - Y^2 = 2a^2$. The $XY$ term coefficient is $-2 \sin 	heta \cos 	heta + 2\sqrt{3}(\cos^2 	heta - \sin^2 	heta) - 2 \sin 	heta \cos 	heta = 0$. This simplifies to $-4 \sin 	heta \cos 	heta + 2\sqrt{3} \cos 2	heta = 0$,which is $-2 \sin 2	heta + 2\sqrt{3} \cos 2	heta = 0$. Thus,$	an 2	heta = \sqrt{3}$,which implies $2	heta = 60^{\circ}$,so $	heta = 30^{\circ}$.

Suppose the axes $X$ and $Y$ are obtained by rotating the axes $x$ and $y$ by an angle $\theta$. If the equation $x^2+2 \sqrt{3} x y-y^2=4 a^2$ is transformed to $X^2-Y^2=2 a^2$ with respect to the $XY$-axes,then $\theta$ is equal to (in $^{\circ}$)

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