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The point to which the origin should be shifted in order to remove the x and y terms in the equation $14 x^{2} - 4 xy + 11 y^{2} - 36 x + 48 y + 41 = 0$ is

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(1,-2)

(-2,1)

(-1,2)

(2,-1)

answer is A.

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Detailed Solution

$\frac{\partial s}{\partial x} = 0, \frac{\partial s}{\partial y} = 0$

$P = (\frac{h f - b g}{a b - h^{2}}, \frac{g h - a f}{a b - h^{2}})$

$a = 14, 2 h = - 4, b = 11, 2 g = - 36, 2 f = 48, c = 41 P = (\frac{150}{150}, \frac{- 300}{150}) = (1, - 2)$

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