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

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Aicharya · Accepted Answer

Given $P(x, y),(-1,0)$ and $(0,2)$ Let $A(-1,0)$ and $B(0,2)$ $ \begin{aligned} & \frac{\mathrm{PA}}{\mathrm{P}(\mathrm{B})}=\frac{\sqrt{2}}{1} \quad 	ext { \{Given\} } \ & \frac{\sqrt{(\mathrm{x}+1)^2+(\mathrm{y}-0)^2}}{\sqrt{(\mathrm{x}-0)^2+(\mathrm{y}-2)^2}}=\frac{\sqrt{2}}{1} \ & \mathrm{x}^2+1+2 \mathrm{x}+\mathrm{y}^2=2 \mathrm{x}^2+2 \mathrm{y}^2+8-8 \mathrm{y} \ & \mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-8 \mathrm{y}+7=0 \ & (\mathrm{x}-1)^2+(\mathrm{y}-4)^2=10 \end{aligned} $

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

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