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(C) The given equation is $f(x, y) = 2x^2+3xy-2y^2-17x+6y+8=0$. To eliminate the linear terms by shifting the origin to $(\alpha, \beta)$,we set the p

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$2\beta$

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(C) The given equation is $f(x, y) = 2x^2+3xy-2y^2-17x+6y+8=0$. To eliminate the linear terms by shifting the origin to $(\alpha, \beta)$,we set the partial derivatives with respect to $x$ and $y$ to zero: $f_x = 4x+3y-17 = 0$ $f_y = 3x-4y+6 = 0$ Solving these equations: Multiply $f_x$ by $4$ and $f_y$ by $3$: $16x+12y-68 = 0$ $9x-12y+18 = 0$ Adding them: $25x = 50 \implies x = 2$. Substituting $x=2$ into $f_x$: $4(2)+3y-17 = 0 \implies 3y = 9 \implies y = 3$. So,the new origin is $(\alpha, \beta) = (2, 3)$. The constant term $c$ in the transformed equation is $f(\alpha, \beta) = 2(2)^2+3(2)(3)-2(3)^2-17(2)+6(3)+8 = 8+18-18-34+18+8 = 0$. Thus,$c = 0$. We need to find $3\alpha+c = 3(2)+0 = 6$. Comparing with the options,since $h$ is the coefficient of $XY$ in the transformed equation,we find $h$ from the original quadratic part $2x^2+3xy-2y^2$. The coefficient of $XY$ is $2h = 3$,so $h = 1.5$. However,checking the options,$2\beta = 2(3) = 6$. Therefore,$3\alpha+c = 2\beta$.

If the transformed equation of the equation $2x^2+3xy-2y^2-17x+6y+8=0$ after translating the coordinate axes to a new origin $(\alpha, \beta)$ is $aX^2+2hXY+bY^2+c=0$,then $3\alpha+c=$

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