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(C) Let the origin be shifted to $(h, k) = (2, 3)$. The transformation equations are $x = X \cos 	heta - Y \sin 	heta + 2$ and $y = X \sin 	heta +

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(C) Let the origin be shifted to $(h, k) = (2, 3)$. The transformation equations are $x = X \cos 	heta - Y \sin 	heta + 2$ and $y = X \sin 	heta + Y \cos 	heta + 3$. Substituting these into the equation $3x^2 + 2xy + 3y^2 - 18x - 22y + 50 = 0$,the $XY$ term must vanish in the transformed equation $4X^2 + 2Y^2 - 1 = 0$. The coefficient of $XY$ in the general quadratic $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ after rotation by $	heta$ is given by $2h' = (b - a) \sin 2	heta + 2h \cos 2	heta$. Here,$a = 3, b = 3, h = 1$. Setting $2h' = 0$,we get $(3 - 3) \sin 2	heta + 2(1) \cos 2	heta = 0$. This implies $2 \cos 2	heta = 0$,so $\cos 2	heta = 0$. Thus,$2	heta = \frac{\pi}{2}$,which gives $	heta = \frac{\pi}{4}$.

By shifting the origin to the point $(2, 3)$ and then rotating the coordinate axes through an angle $\theta$ in the counter-clockwise direction,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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