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(D) Substituting $x=X+\frac{3}{2}$ and $y=Y+\frac{3}{2}$ into the equation $32 x^2+8 x y+32 y^2-108 x-108 y+99=0$: $32(X+\frac{3}{2})^2 + 8(X+\frac{3}

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$32 X^2+8 X Y+32 Y^2-63=0$

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(D) Substituting $x=X+\frac{3}{2}$ and $y=Y+\frac{3}{2}$ into the equation $32 x^2+8 x y+32 y^2-108 x-108 y+99=0$: $32(X+\frac{3}{2})^2 + 8(X+\frac{3}{2})(Y+\frac{3}{2}) + 32(Y+\frac{3}{2})^2 - 108(X+\frac{3}{2}) - 108(Y+\frac{3}{2}) + 99 = 0$ Expanding the terms: $32(X^2 + 3X + \frac{9}{4}) + 8(XY + \frac{3}{2}X + \frac{3}{2}Y + \frac{9}{4}) + 32(Y^2 + 3Y + \frac{9}{4}) - 108X - 162 - 108Y - 162 + 99 = 0$ $32X^2 + 96X + 72 + 8XY + 12X + 12Y + 18 + 32Y^2 + 96Y + 72 - 108X - 108Y - 162 - 162 + 99 = 0$ Combining like terms: $32X^2 + 32Y^2 + 8XY + (96+12-108)X + (96+12-108)Y + (72+18+72-162-162+99) = 0$ $32X^2 + 8XY + 32Y^2 - 63 = 0$

When the origin is shifted to the point $\left(\frac{3}{2}, \frac{3}{2}\right)$ by the translation of coordinate axes,then the transformed equation of $32 x^2+8 x y+32 y^2-108 x-108 y+99=0$ is

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