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Question

(A) Let the axes be rotated through an angle $	heta$. The transformation equations are $x = X \cos 	heta - Y \sin 	heta$ and $y = X \sin 	heta + Y

Vedclass · Accepted Answer

$(2+\sqrt{3})x^2+(2-\sqrt{3})y^2+2xy=4$

Vedclass · Answer

(A) Let the axes be rotated through an angle $	heta$. The transformation equations are $x = X \cos 	heta - Y \sin 	heta$ and $y = X \sin 	heta + Y \cos 	heta$.For the equation $3x^2 + 2\sqrt{3}xy + y^2 = 0$,the coefficient of $XY$ is $2(A-B)\sin 	heta \cos 	heta + 2H(\cos^2 	heta - \sin^2 	heta)$.Here $A=3, H=\sqrt{3}, B=1$. Setting the new $XY$ coefficient to $0$ gives $2(3-1)\sin 	heta \cos 	heta + 2\sqrt{3}\cos 2	heta = 0$,which simplifies to $2\sin 2	heta + 2\sqrt{3}\cos 2	heta = 0$.This implies $	an 2	heta = -\sqrt{3}$,so $2	heta = 120^{\circ}$ or $	heta = 60^{\circ}$.Substituting $	heta = 60^{\circ}$ into $x^2 + y^2 + 2xy = 2$,we use $x+y = X(\cos 	heta + \sin 	heta) + Y(\cos 	heta - \sin 	heta) = X(\frac{1+\sqrt{3}}{2}) + Y(\frac{1-\sqrt{3}}{2})$.Since $x^2+y^2+2xy = (x+y)^2 = 2$,we have $(x+y)^2 = 2$.Substituting the values,we get the transformed equation as $(2+\sqrt{3})X^2 + (2-\sqrt{3})Y^2 + 2XY = 4$.

Suppose the axes are to be rotated through an angle $\theta$ so as to remove the $xy$ term from the equation $3x^2+2\sqrt{3}xy+y^2=0$. Then in the new coordinate system,the equation $x^2+y^2+2xy=2$ is transformed to:

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