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(D) Given equation: $2x^2 - 3y^2 + 4xy + 4x + 4y - 14 = 0$. Let the origin be shifted to $(h, k)$,so $x = X + h$ and $y = Y + k$. Substituting these i

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$x^2 + y^2 - 3xy - 2x + 7y + 4 = 0$

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(D) Given equation: $2x^2 - 3y^2 + 4xy + 4x + 4y - 14 = 0$. Let the origin be shifted to $(h, k)$,so $x = X + h$ and $y = Y + k$. Substituting these into the equation: $2(X+h)^2 - 3(Y+k)^2 + 4(X+h)(Y+k) + 4(X+h) + 4(Y+k) - 14 = 0$. Expanding the terms: $2X^2 - 3Y^2 + 4XY + (4h + 4k + 4)X + (4h - 6k + 4)Y + (2h^2 - 3k^2 + 4hk + 4h + 4k - 14) = 0$. To remove the first-degree terms,set the coefficients of $X$ and $Y$ to zero: $4h + 4k + 4 = 0 \Rightarrow h + k = -1$ $4h - 6k + 4 = 0 \Rightarrow 2h - 3k = -2$ Solving these equations: $h = -1, k = 0$. Thus,the transformation is $x = X - 1$ and $y = Y$. Now,substitute these into $x^2 + y^2 - 3xy + 4y + 3 = 0$: $(X - 1)^2 + Y^2 - 3(X - 1)Y + 4Y + 3 = 0$ $X^2 - 2X + 1 + Y^2 - 3XY + 3Y + 4Y + 3 = 0$ $X^2 + Y^2 - 3XY - 2X + 7Y + 4 = 0$.

If the origin is shifted to remove the first degree terms from the equation $2x^2 - 3y^2 + 4xy + 4x + 4y - 14 = 0$,then with respect to this new coordinate system,the transformed equation of $x^2 + y^2 - 3xy + 4y + 3 = 0$ is

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