When the origin is shifted to the point (2,3) | vedclass

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(B) Step $1$: Shift the origin to $(2,3)$. Substitute $x = X+2$ and $y = Y+3$ into the equation $3x^2+2xy+3y^2-18x-22y+50=0$.$3(X+2)^2+2(X+2)(Y+3)+3(Y

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$(6+\sqrt{3}) x^2-2 x y+(6-\sqrt{3}) y^2-2=0$

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(B) Step $1$: Shift the origin to $(2,3)$. Substitute $x = X+2$ and $y = Y+3$ into the equation $3x^2+2xy+3y^2-18x-22y+50=0$.$3(X+2)^2+2(X+2)(Y+3)+3(Y+3)^2-18(X+2)-22(Y+3)+50=0$$3(X^2+4X+4)+2(XY+3X+2Y+6)+3(Y^2+6Y+9)-18X-36-22Y-66+50=0$$3X^2+12X+12+2XY+6X+4Y+12+3Y^2+18Y+27-18X-36-22Y-66+50=0$Simplifying,we get $3X^2+2XY+3Y^2-1=0$.Step $2$: Rotate the axes by $	heta = \frac{\pi}{3}$ counter-clockwise. The transformation equations are:$X = x' \cos\frac{\pi}{3} - y' \sin\frac{\pi}{3} = \frac{x' - \sqrt{3}y'}{2}$$Y = x' \sin\frac{\pi}{3} + y' \cos\frac{\pi}{3} = \frac{\sqrt{3}x' + y'}{2}$Substitute these into $3X^2+2XY+3Y^2-1=0$:$3(\frac{x'-\sqrt{3}y'}{2})^2 + 2(\frac{x'-\sqrt{3}y'}{2})(\frac{\sqrt{3}x'+y'}{2}) + 3(\frac{\sqrt{3}x'+y'}{2})^2 - 1 = 0$$\frac{3}{4}(x'^2+3y'^2-2\sqrt{3}x'y') + \frac{2}{4}(\sqrt{3}x'^2+x'y'-3x'y'-\sqrt{3}y'^2) + \frac{3}{4}(3x'^2+y'^2+2\sqrt{3}x'y') - 1 = 0$Multiplying by $4$:$3(x'^2+3y'^2-2\sqrt{3}x'y') + 2(\sqrt{3}x'^2-2x'y'-\sqrt{3}y'^2) + 3(3x'^2+y'^2+2\sqrt{3}x'y') - 4 = 0$$(3+2\sqrt{3}+9)x'^2 + (-6\sqrt{3}-4+6\sqrt{3})x'y' + (9-2\sqrt{3}+3)y'^2 - 4 = 0$$(12+2\sqrt{3})x'^2 - 4x'y' + (12-2\sqrt{3})y'^2 - 4 = 0$Dividing by $2$:$(6+\sqrt{3})x^2 - 2xy + (6-\sqrt{3})y^2 - 2 = 0$.

When the origin is shifted to the point $(2,3)$ and then the coordinate axes are rotated through an angle $\frac{\pi}{3}$ in the counter-clockwise sense,then the transformed equation of $3 x^2+2 x y+3 y^2-18 x-22 y+50=0$ is

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