By shifting the origin to the point (2,3) thr | vedclass

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(A) Let the original coordinates be $(x, y)$ and the new coordinates be $(X, Y)$.The transformation equations are $x = X + h$ and $y = Y + k$,where $(

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$-1$

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(A) Let the original coordinates be $(x, y)$ and the new coordinates be $(X, Y)$.The transformation equations are $x = X + h$ and $y = Y + k$,where $(h, k) = (2, 3)$.So,$x = X + 2$ and $y = Y + 3$.Substituting these into the original equation $x^2 + 3xy - 2y^2 + 4x - y - 20 = 0$:$(X+2)^2 + 3(X+2)(Y+3) - 2(Y+3)^2 + 4(X+2) - (Y+3) - 20 = 0$$(X^2 + 4X + 4) + 3(XY + 3X + 2Y + 6) - 2(Y^2 + 6Y + 9) + 4X + 8 - Y - 3 - 20 = 0$$X^2 + 4X + 4 + 3XY + 9X + 6Y + 18 - 2Y^2 - 12Y - 18 + 4X - Y - 3 - 20 = 0$Grouping the terms:$X^2 + 3XY - 2Y^2 + (4 + 9 + 4)X + (6 - 12 - 1)Y + (4 + 18 - 18 - 3 - 20) = 0$$X^2 + 3XY - 2Y^2 + 17X - 7Y - 19 = 0$Comparing this with $AX^2 + BXY + CY^2 + DX + EY + F = 0$,we get $D = 17$,$E = -7$,and $F = -19$.Therefore,$D + E + F = 17 - 7 - 19 = -9$.Wait,re-calculating the constant term: $4 + 18 - 18 - 3 - 20 = 4 - 23 = -19$.Re-calculating $D$: $4 + 9 + 4 = 17$.Re-calculating $E$: $6 - 12 - 1 = -7$.Sum $D+E+F = 17 - 7 - 19 = -9$. Since $-9$ is not in the options,let us re-check the calculation.$x^2+3xy-2y^2+4x-y-20=0$ at $(2,3)$:$x^2+4x+4 + 3(xy+3x+2y+6) - 2(y^2+6y+9) + 4x+8 - y-3 - 20 = 0$$x^2+3xy-2y^2 + (4+9+4)x + (6-12-1)y + (4+18-18+8-3-20) = 0$$x^2+3xy-2y^2 + 17x - 7y - 11 = 0$.$D+E+F = 17 - 7 - 11 = -1$.

By shifting the origin to the point $(2,3)$ through translation of axes,if the equation of the curve $x^2+3xy-2y^2+4x-y-20=0$ is transformed to the form $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$,then $D+E+F=$

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