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Without changing the direction of coordinates axes, origin is transferred to (\alpha, \beta) so that the linear terms in the equation \mathrm{x^{2}+y^{2}+2 x-4 y+6=0} are eliminated. The point (\alpha, \beta) is

Option: 1

(-1,2)


Option: 2

(1,-2)


Option: 3

(1,2)


Option: 4

(-1,-2)


Answers (1)

best_answer

The given equation is \mathrm{x^{2}+y^{2}+2 x-4 y+6=0 \quad \ldots . .(1)}

Putting \mathrm{x=x^{\prime}+\alpha \: and\: y=y^{\prime}+\beta\, in (1) we \: get x^{\prime 2}+y^{\prime 2}+x^{\prime}(2 \alpha+2)+y^{\prime}(2 \beta-4)+\left(\alpha^{2}+\beta^{2}+2 \alpha-\right.4 \beta+6)=0}.
To eliminate linear term, we should have \mathrm{2 \alpha+2=0} and \mathrm{2 \beta-4=0}

\mathrm{\Rightarrow \alpha=-1\, and \, \beta=2}
\mathrm{\Rightarrow(\alpha, \beta) \equiv(-1,2)}
 

Posted by

sudhir.kumar

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