By hypothesis,
$\left| OP\right| = 2 \left |PM\right|$
$\Rightarrow \sqrt{x_{1}^{2}+y_{1}^{2}}=2\left| x_{1}\right|$
On squaring both sides, we get
$x_{1}^{2}+y_{1}^{2}=4 x_{1}^{2} \Rightarrow 3 x_{1}^{2}-y_{1}^{2}=0$
$\therefore $ Locus of the point is $3 x^{2}-y^{2}=0$