The origin is shifted to the point (2,3) by t | vedclass

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(A) Given equation: $3x^2+2xy+3y^2-18x-22y+50=0$.Shifting origin to $(2,3)$,substitute $x=X+2, y=Y+3$:$3(X+2)^2+2(X+2)(Y+3)+3(Y+3)^2-18(X+2)-22(Y+3)+5

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$\frac{\pi}{4}$

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(A) Given equation: $3x^2+2xy+3y^2-18x-22y+50=0$.Shifting origin to $(2,3)$,substitute $x=X+2, y=Y+3$:$3(X+2)^2+2(X+2)(Y+3)+3(Y+3)^2-18(X+2)-22(Y+3)+50=0$.Simplifying,we get $3X^2+2XY+3Y^2-1=0$.Now,rotating axes by angle $	heta$,substitute $X=x'\cos	heta-y'\sin	heta$ and $Y=x'\sin	heta+y'\cos	heta$:$3(x'\cos	heta-y'\sin	heta)^2+2(x'\cos	heta-y'\sin	heta)(x'\sin	heta+y'\cos	heta)+3(x'\sin	heta+y'\cos	heta)^2-1=0$.Expanding and collecting terms:$(3\cos^2	heta+3\sin^2	heta+2\sin	heta\cos	heta)x'^2 + (3\sin^2	heta+3\cos^2	heta-2\sin	heta\cos	heta)y'^2 + (2\cos^2	heta-2\sin^2	heta)x'y' - 1 = 0$.$(3+\sin 2	heta)x'^2 + (3-\sin 2	heta)y'^2 + (2\cos 2	heta)x'y' - 1 = 0$.Comparing with $4x^2+2y^2-1=0$,the coefficient of $x'y'$ must be $0$:$2\cos 2	heta = 0$ $\Rightarrow 2	heta = \frac{\pi}{2}$ $\Rightarrow 	heta = \frac{\pi}{4}$.

The origin is shifted to the point $(2,3)$ by translation of axes and then the coordinate axes are rotated about the origin through an angle $\theta$ in the counter-clockwise sense. Due to this,if the equation $3x^2+2xy+3y^2-18x-22y+50=0$ is transformed to $4x^2+2y^2-1=0$,then the angle $\theta=$

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