The transformed equation of 3x^2+4xy+y^2-8x-4 | vedclass

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(A) The given equation is $3x^2+4xy+y^2-8x-4y-4=0$. To eliminate the linear terms,we shift the origin to $(h, k)$. Let $x = X+h$ and $y = Y+k$. Substi

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(A) The given equation is $3x^2+4xy+y^2-8x-4y-4=0$. To eliminate the linear terms,we shift the origin to $(h, k)$. Let $x = X+h$ and $y = Y+k$. Substituting these into the equation: $3(X+h)^2 + 4(X+h)(Y+k) + (Y+k)^2 - 8(X+h) - 4(Y+k) - 4 = 0$. Expanding this,we get: $3X^2 + 4XY + Y^2 + X(6h+4k-8) + Y(4h+2k-4) + (3h^2+4hk+k^2-8h-4k-4) = 0$. For the linear terms to vanish,we set the coefficients of $X$ and $Y$ to zero: $6h+4k-8 = 0 \Rightarrow 3h+2k=4$ $4h+2k-4 = 0 \Rightarrow 2h+k=2$ Solving these,we get $h=0$ and $k=2$. Substituting $h=0, k=2$ into the constant term: $c = 3(0)^2 + 4(0)(2) + (2)^2 - 8(0) - 4(2) - 4 = 4 - 8 - 4 = -8$. Thus,the transformed equation is $f(X, Y) = 3X^2 + 4XY + Y^2 - 8 = 0$. Therefore,$f(1,1) = 3(1)^2 + 4(1)(1) + (1)^2 - 8 = 3 + 4 + 1 - 8 = 0$.

The transformed equation of $3x^2+4xy+y^2-8x-4y-4=0$ is $f(X, Y)=aX^2+2hXY+bY^2+c=0$ when the origin is shifted to a new point by the translation of axes. Then $f(1,1)=$

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