To convert the equation 2x^2 + 4xy + 5y^2 - 4 | vedclass

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(B) Let the given equation be $f(x, y) = 2x^2 + 4xy + 5y^2 - 4x - 22y + 29 = 0$. \ To shift the origin to $(h, k)$,we substitute $x = X + h$ and $y =

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$(-2, 3)$

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(B) Let the given equation be $f(x, y) = 2x^2 + 4xy + 5y^2 - 4x - 22y + 29 = 0$. \ To shift the origin to $(h, k)$,we substitute $x = X + h$ and $y = Y + k$. \ The equation becomes $2(X+h)^2 + 4(X+h)(Y+k) + 5(Y+k)^2 - 4(X+h) - 22(Y+k) + 29 = 0$. \ Expanding this,the linear terms in $X$ and $Y$ must vanish for the equation to be homogeneous. \ The partial derivatives with respect to $x$ and $y$ are: \ $f_x = 4x + 4y - 4 = 0 \implies x + y = 1$ \ $f_y = 4x + 10y - 22 = 0 \implies 2x + 5y = 11$ \ Solving these equations: \ From the first,$x = 1 - y$. 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 must be shifted to $(-2, 3)$.

To convert the equation $2x^2 + 4xy + 5y^2 - 4x - 22y + 29 = 0$ to homogeneous form,the origin is shifted to the point:

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