If the equation 3x^2 + 4y^2 - xy + k = 0 is t | vedclass

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(B) Let the origin be shifted to $(\alpha, \beta)$. The transformation equations are $x = X + \alpha$ and $y = Y + \beta$.Substituting these into the

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$6$

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(B) Let the origin be shifted to $(\alpha, \beta)$. The transformation equations are $x = X + \alpha$ and $y = Y + \beta$.Substituting these into the original equation $3x^2 + 4y^2 - xy - 5x - 7y + 2 = 0$:$3(X + \alpha)^2 + 4(Y + \beta)^2 - (X + \alpha)(Y + \beta) - 5(X + \alpha) - 7(Y + \beta) + 2 = 0$Expanding this,the coefficients of $X$ and $Y$ must be zero for the equation to be of the form $3X^2 + 4Y^2 - XY + k = 0$.The coefficient of $X$ is $6\alpha - \beta - 5 = 0$.The coefficient of $Y$ is $8\beta - \alpha - 7 = 0$.Solving these linear equations: $\alpha = 1, \beta = 1$.Substituting $\alpha = 1, \beta = 1$ into the constant term:$k = 3(1)^2 + 4(1)^2 - (1)(1) - 5(1) - 7(1) + 2 = 3 + 4 - 1 - 5 - 7 + 2 = -4$.Thus,$\alpha + \beta - k = 1 + 1 - (-4) = 6$.

If the equation $3x^2 + 4y^2 - xy + k = 0$ is the transformed equation of $3x^2 + 4y^2 - xy - 5x - 7y + 2 = 0$ after shifting the origin to the point $(\alpha, \beta)$ by the translation of axes,then $\alpha + \beta - k =$

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