If 2x^2+xy-6y^2+k=0 is the transformed equati | vedclass

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(B) Let the original equation be $f(x, y) = 2x^2+xy-6y^2-13x+9y+15=0$. When the origin is shifted to $(a, b)$,we substitute $x = X+a$ and $y = Y+b$. T

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(B) Let the original equation be $f(x, y) = 2x^2+xy-6y^2-13x+9y+15=0$. When the origin is shifted to $(a, b)$,we substitute $x = X+a$ and $y = Y+b$. The equation becomes $2(X+a)^2 + (X+a)(Y+b) - 6(Y+b)^2 - 13(X+a) + 9(Y+b) + 15 = 0$. Expanding this,the linear terms in $X$ and $Y$ must vanish for the equation to take the form $2X^2+XY-6Y^2+k=0$. The partial derivatives $\frac{\partial f}{\partial x} = 4x+y-13 = 0$ and $\frac{\partial f}{\partial y} = x-12y+9 = 0$ give the center $(a, b)$. Solving $4a+b=13$ and $a-12b=-9$,we get $a=3$ and $b=1$. Substituting $x=X+3$ and $y=Y+1$ into the original equation: $2(X+3)^2 + (X+3)(Y+1) - 6(Y+1)^2 - 13(X+3) + 9(Y+1) + 15 = 0$. $2(X^2+6X+9) + (XY+X+3Y+3) - 6(Y^2+2Y+1) - 13X-39 + 9Y+9 + 15 = 0$. $2X^2+12X+18 + XY+X+3Y+3 - 6Y^2-12Y-6 - 13X+9Y+9+15 = 0$. Grouping terms: $2X^2+XY-6Y^2 + (12+1-13)X + (3-12+9)Y + (18+3-6-39+9+15) = 0$. $2X^2+XY-6Y^2 + 0X + 0Y + 0 = 0$. Thus,$k=0$.

If $2x^2+xy-6y^2+k=0$ is the transformed equation of $2x^2+xy-6y^2-13x+9y+15=0$ when the origin is shifted to the point $(a, b)$ by translation of axes,then $k=$

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