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(C) Given equation: $ax^2-2xy+by^2-2x+4y+1=0$. Shift origin to $(0, k)$,so $x=X$ and $y=Y+k$. Substituting these in the equation: $aX^2-2X(Y+k)+b(Y+k)

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$3$

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(C) Given equation: $ax^2-2xy+by^2-2x+4y+1=0$. Shift origin to $(0, k)$,so $x=X$ and $y=Y+k$. Substituting these in the equation: $aX^2-2X(Y+k)+b(Y+k)^2-2X+4(Y+k)+1=0$. Expanding: $aX^2-2XY-2kX+bY^2+bk^2+2bkY-2X+4Y+4k+1=0$. Grouping terms: $aX^2-2XY+bY^2-2X(k+1)+2Y(bk+2)+bk^2+4k+1=0$. To remove first degree terms,coefficients of $X$ and $Y$ must be zero: $k+1=0 \Rightarrow k=-1$ and $bk+2=0$ $\Rightarrow -b+2=0$ $\Rightarrow b=2$. Now,rotate axes by $	heta = \frac{1}{2} 	an^{-1}(2)$,so $	an(2	heta) = 2$. The $xy$ term in the rotated equation is removed if $	an(2	heta) = \frac{2h}{a-b}$,where $h=-1$. Thus,$	an(2	heta) = \frac{2(-1)}{a-b} = \frac{-2}{a-b} = \frac{2}{b-a}$. Given $	an(2	heta) = 2$,we have $\frac{2}{b-a} = 2 \Rightarrow b-a = 1$. Since $b=2$,$2-a=1 \Rightarrow a=1$. Therefore,$a+b = 1+2 = 3$.

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If $O$ is the origin and the coordinates of any two points $Q_1$ and $Q_2$ are $(x_1, y_1)$ and $(x_2, y_2)$ respectively,then $OQ_1 \cdot OQ_2 \cos \angle Q_1OQ_2 = $

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