Let P(alpha, beta) be a variable point which | vedclass

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(C) The locus of $P(x, y)$ is given by $\frac{PA}{PB} = 2$,which implies $PA^2 = 4PB^2$. $(x-1)^2 + y^2 = 4(x^2 + (y+1)^2)$. $x^2 - 2x + 1 + y^2 = 4x^

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(C) The locus of $P(x, y)$ is given by $\frac{PA}{PB} = 2$,which implies $PA^2 = 4PB^2$. $(x-1)^2 + y^2 = 4(x^2 + (y+1)^2)$. $x^2 - 2x + 1 + y^2 = 4x^2 + 4y^2 + 8y + 4$. $3x^2 + 3y^2 + 2x + 8y + 3 = 0$. $x^2 + y^2 + \frac{2}{3}x + \frac{8}{3}y + 1 = 0$. The center is $C(-\frac{1}{3}, -\frac{4}{3})$ and the radius $r = \sqrt{(-\frac{1}{3})^2 + (-\frac{4}{3})^2 - 1} = \sqrt{\frac{1}{9} + \frac{16}{9} - 1} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$. Let $\alpha = -\frac{1}{3} + r \cos 	heta$ and $\beta = -\frac{4}{3} + r \sin 	heta$. Then $\alpha + \beta = -\frac{5}{3} + r(\cos 	heta + \sin 	heta) = -\frac{5}{3} + r\sqrt{2} \sin(	heta + \frac{\pi}{4})$. Since $r = \frac{2\sqrt{2}}{3}$,$r\sqrt{2} = \frac{4}{3}$. Max value $M = -\frac{5}{3} + \frac{4}{3} = -\frac{1}{3}$. Min value $m = -\frac{5}{3} - \frac{4}{3} = -\frac{9}{3} = -3$. Thus,$\frac{M}{m} = \frac{-1/3}{-3} = \frac{1}{9}$. $[\frac{M}{m}] = [\frac{1}{9}] = 0$.

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