A point P(x, y) is such that its distances fr | vedclass

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Question

(A) Let $A = (-1, 0)$ and $B = (0, 2)$.Given the ratio of distances $\frac{PA}{PB} = \frac{\sqrt{2}}{1}$.Squaring both sides,we get $\frac{PA^2}{PB^2}

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$(x-1)^2+(y-4)^2=10$

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(A) Let $A = (-1, 0)$ and $B = (0, 2)$.Given the ratio of distances $\frac{PA}{PB} = \frac{\sqrt{2}}{1}$.Squaring both sides,we get $\frac{PA^2}{PB^2} = 2$.$PA^2 = (x+1)^2 + (y-0)^2 = x^2 + 2x + 1 + y^2$.$PB^2 = (x-0)^2 + (y-2)^2 = x^2 + y^2 - 4y + 4$.Substituting into the ratio equation: $x^2 + 2x + 1 + y^2 = 2(x^2 + y^2 - 4y + 4)$.$x^2 + 2x + 1 + y^2 = 2x^2 + 2y^2 - 8y + 8$.Rearranging terms: $x^2 + y^2 - 2x - 8y + 7 = 0$.Completing the square: $(x^2 - 2x + 1) + (y^2 - 8y + 16) = -7 + 1 + 16$.$(x-1)^2 + (y-4)^2 = 10$.

$A$ point $P(x, y)$ is such that its distances from $(-1, 0)$ and $(0, 2)$ are in the ratio $\sqrt{2} : 1$. Then the locus of $P$ is

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