The locus of a point which moves such that it | vedclass

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(B) Let the point be $P(h, k)$.$P_{1}$ is the length of the perpendicular from $P$ to the $x$-axis $(y=0)$,so $P_{1} = |k|$.$P_{2}$ is the length of t

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$2x^{2}-4xy+y^{2}=0$

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(B) Let the point be $P(h, k)$.$P_{1}$ is the length of the perpendicular from $P$ to the $x$-axis $(y=0)$,so $P_{1} = |k|$.$P_{2}$ is the length of the perpendicular from $P$ to the line $x-y=0$,so $P_{2} = \frac{|h-k|}{\sqrt{1^{2}+(-1)^{2}}} = \frac{|h-k|}{\sqrt{2}}$.Given that $P_{1} = 2P_{2}$,we have:$|k| = 2 \cdot \frac{|h-k|}{\sqrt{2}}$$|k| = \sqrt{2} |h-k|$Squaring both sides:$k^{2} = 2(h-k)^{2}$$k^{2} = 2(h^{2} + k^{2} - 2hk)$$k^{2} = 2h^{2} + 2k^{2} - 4hk$$2h^{2} - 4hk + k^{2} = 0$Replacing $(h, k)$ with $(x, y)$,the locus is:$2x^{2} - 4xy + y^{2} = 0$

The locus of a point which moves such that its distance from the $x$-axis is twice its distance from the line $x-y=0$ is

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