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

• A

  $x^{2}+4 x y-y^{2}=0$
• Correct answer
  B

  $2 x^{2}-4 x y+y^{2}=0$
• C

  $x^{2}-4 x y+y^{2}=0$
• D

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

Solution

$P_{1}=$ length of perpendicular from $P$ to $x$-axis. $P_{2}=$ length of perpendicular from $P$ to $y=x$ line. $$ \begin{aligned} P_{1} &=2 P_{2} \\ |k| &=2 \cdot \frac{|h-k|}{\sqrt{2}} \end{aligned} $$ Squaring on both sides, $$ k^{2}=2(h-k)^{2} $$

$$ \begin{array}{r} k^{2}=2 h^{2}+2 k^{2}-4 h k \\ \Rightarrow \quad 2 h^{2}-4 h k+k^{2}=0 \end{array} $$ So, the locus of a point $P$ is, $$ 2 x^{2}-4 x y+y^{2}=0 $$

Hence, the correct answer is option: 'B'.