Find the center of the conic 14x^2 - 4xy + 11 | vedclass

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Question

(A) Let $f(x, y) = 14x^2 - 4xy + 11y^2 - 44x - 58y + 71 = 0$.Taking partial derivatives with respect to $x$ and $y$:$\frac{\partial f}{\partial x} = 2

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$(2, 3)$

Vedclass · Answer

(A) Let $f(x, y) = 14x^2 - 4xy + 11y^2 - 44x - 58y + 71 = 0$.Taking partial derivatives with respect to $x$ and $y$:$\frac{\partial f}{\partial x} = 28x - 4y - 44$ and $\frac{\partial f}{\partial y} = -4x + 22y - 58$.For the center,set $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$:$28x - 4y - 44 = 0 \implies 7x - y = 11$ (Equation $I$)$-4x + 22y - 58 = 0 \implies -2x + 11y = 29$ (Equation $II$)Solving the system of linear equations:From $I$,$y = 7x - 11$.Substitute into $II$: $-2x + 11(7x - 11) = 29 \implies -2x + 77x - 121 = 29 \implies 75x = 150 \implies x = 2$.Then $y = 7(2) - 11 = 3$.Therefore,the center is $(2, 3)$.

Find the center of the conic $14x^2 - 4xy + 11y^2 - 44x - 58y + 71 = 0$.

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