In order to eliminate the first degree terms | vedclass

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(C) The given equation is $2x^2+4xy+5y^2-4x-22y+7=0$. Comparing this with the general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=2,

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$(-2, 3)$

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(C) The given equation is $2x^2+4xy+5y^2-4x-22y+7=0$. Comparing this with the general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=2, h=2, b=5, g=-2, f=-11, c=7$. To eliminate the first-degree terms,the origin $(0,0)$ must be shifted to the point $(h', k')$ given by the intersection of the partial derivatives $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$. $\frac{\partial f}{\partial x} = 4x+4y-4 = 0 \implies x+y=1$ $\frac{\partial f}{\partial y} = 4x+10y-22 = 0 \implies 2x+5y=11$ Solving these equations: From the first equation,$x = 1-y$. Substituting into the second: $2(1-y)+5y=11 \implies 2-2y+5y=11 \implies 3y=9 \implies y=3$. Then $x = 1-3 = -2$. Thus,the origin is shifted to $(-2, 3)$.

In order to eliminate the first degree terms from the equation $2x^2+4xy+5y^2-4x-22y+7=0$,the point to which the origin is to be shifted is

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