When the origin is shifted to the point (h, k | vedclass

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(A) The original equation is $2x^2 - xy + y^2 + 2x + 3y + 1 = 0$. To shift the origin to $(h, k)$,we substitute $x = X + h$ and $y = Y + k$. The new e

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-$4$

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(A) The original equation is $2x^2 - xy + y^2 + 2x + 3y + 1 = 0$. To shift the origin to $(h, k)$,we substitute $x = X + h$ and $y = Y + k$. The new equation is $2(X+h)^2 - (X+h)(Y+k) + (Y+k)^2 + 2(X+h) + 3(Y+k) + 1 = 0$. Expanding this,the linear terms in $X$ and $Y$ are $(4h - k + 2)X + (-h + 2k + 3)Y = 0$. For the origin to be the new center,these coefficients must be zero: $4h - k = -2$ and $-h + 2k = -3$. Solving these,we get $h = -1$ and $k = -2$. Thus,$h + k = -3$. The equation becomes $2X^2 - XY + Y^2 + C' = 0$. To eliminate the $XY$ term by rotation,we use $	an 2	heta = \frac{2H}{A-B}$,where the equation is $AX^2 + 2HXY + BY^2 = 0$. Here $A = 2, H = -1/2, B = 1$. So,$	an 2	heta = \frac{2(-1/2)}{2 - 1} = \frac{-1}{1} = -1$. Therefore,$h + k + 	an 2	heta = -3 + (-1) = -4$.

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