Find the centre of the conic section represen | vedclass

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(A) The given equation is of the form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Comparing this with $14x^2 - 4xy + 11y^2 - 44x - 58y + 71 = 0$,we get:$

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$(2, 3)$

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(A) The given equation is of the form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Comparing this with $14x^2 - 4xy + 11y^2 - 44x - 58y + 71 = 0$,we get:$a = 14, h = -2, b = 11, g = -22, f = -29, c = 71$.The centre $(x, y)$ of the conic is given by the partial derivatives $\frac{\partial}{\partial x} = 0$ and $\frac{\partial}{\partial y} = 0$.$\frac{\partial}{\partial x} = 28x - 4y - 44 = 0 \implies 7x - y = 11$  $(1)$$\frac{\partial}{\partial y} = -4x + 22y - 58 = 0 \implies -2x + 11y = 29$  $(2)$Multiplying $(1)$ by $11$,we get $77x - 11y = 121$.Adding this to $(2)$,we get $75x = 150$,so $x = 2$.Substituting $x = 2$ in $(1)$,$7(2) - y = 11 \implies 14 - y = 11 \implies y = 3$.Thus,the centre is $(2, 3)$.

Find the centre of the conic section represented by the equation $14x^2 - 4xy + 11y^2 - 44x - 58y + 71 = 0$.

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