The centre of the conic section $14 x^{2} - 4 x y + 11 y^{2} - 44 x - 58 y + 71 = 0$ is

(2, 3)

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(2, - 3)

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(- 2, 3)

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(- 2, - 3)

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Solution

The correct option is A $(2, 3)$
Let $S \equiv 14 x^{2} - 4 x y + 11 y^{2} - 44 x - 58 y + 71$
To know centre of this hyperbola, $\frac{\partial S}{\partial x} = 0$ and $\frac{\partial S}{\partial y} = 0$
$\Rightarrow \frac{\partial S}{\partial x} = 28 x - 4 y - 44 = 0 \Rightarrow 7 x - y - 11 = 0 . . (1)$
and $\frac{\partial S}{\partial y} = - 4 x + 22 y - 58 = 0 \Rightarrow 2 x - 11 y + 29 = 0 . . (2)$
Solving $(1)$ and $(2)$ we get required centre $(2, 3)$
Hence, option 'A' is correct.

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The centre of the conic section 14x2−4xy+11y2−44x−58y+71=0 is

The centre of the conic section $14 x^{2} - 4 x y + 11 y^{2} - 44 x - 58 y + 71 = 0$ is