If all chords of the curve 2x^2 - y^2 + 3x + | vedclass

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(A) The equation of the curve is $2x^2 - y^2 + 3x + 2y = 0$. Let the equation of the chord be $y = mx + c$,which can be written as $\frac{y - mx}{c} =

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$(-3, -2)$

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(A) The equation of the curve is $2x^2 - y^2 + 3x + 2y = 0$. Let the equation of the chord be $y = mx + c$,which can be written as $\frac{y - mx}{c} = 1$. Substituting this into the homogeneous form of the curve equation: $2x^2 - y^2 + (3x + 2y)(\frac{y - mx}{c}) = 0$. Multiplying by $c$: $2cx^2 - cy^2 + 3xy - 3mx^2 + 2y^2 - 2mxy = 0$. Grouping the terms: $(2c - 3m)x^2 + (2 - c)y^2 + (3 - 2m)xy = 0$. Since the chord subtends a right angle at the origin,the sum of the coefficients of $x^2$ and $y^2$ must be zero: $(2c - 3m) + (2 - c) = 0$. $c - 3m + 2 = 0$. Substituting $c = 3m - 2$ into the chord equation $y = mx + c$: $y = mx + 3m - 2$. $y + 2 = m(x + 3)$. This equation represents a family of lines passing through the fixed point $(-3, -2)$. Thus,$(\alpha, \beta) = (-3, -2)$.

If all chords of the curve $2x^2 - y^2 + 3x + 2y = 0$,which subtend a right angle at the origin,always pass through a fixed point $(\alpha, \beta)$,then $(\alpha, \beta) =$

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