The point to which the origin is to be shifte | vedclass

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(C) The given equation is $2x^2+4xy-6y^2+2x+8y+1=0$. 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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$\left(\frac{-7}{8}, \frac{3}{8}\right)$

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(C) The given equation is $2x^2+4xy-6y^2+2x+8y+1=0$. Comparing this with the general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=2, h=2, b=-6, g=1, f=4, c=1$. To remove the first-degree terms,the origin must be shifted to the point $(h_0, k_0)$ given by the intersection of the partial derivatives $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$. These are $4x+4y+2=0$ and $4x-12y+8=0$. Solving these equations: $2x+2y=-1$ $2x-6y=-4$ Subtracting the two equations: $8y=3 \implies y=\frac{3}{8}$. Substituting $y=\frac{3}{8}$ into $2x+2y=-1$: $2x+2(\frac{3}{8})=-1 \implies 2x+\frac{3}{4}=-1 \implies 2x=-\frac{7}{4} \implies x=-\frac{7}{8}$. Thus,the required point is $\left(-\frac{7}{8}, \frac{3}{8}\right)$.

The point to which the origin is to be shifted to remove the first degree terms from the equation $2x^2+4xy-6y^2+2x+8y+1=0$ is

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