Question

Find locus of a point so that its distance from the axis of x is always one half its distance from the origin.

Answer 1

Let the point be $P(h,k)$

$PO$ be the distance from the origin

$PO=\sqrt{{(h-0)}^2+{(k-0)}^2}=\sqrt{h^2+k^2}$

$PA$ be the distance from $x$ axis

Equation of $x$ axis is $y=0$

$PA=\frac{0\left(h\right)+1\left(k\right)+0}{\sqrt{0^2+1^2}}=k$

Given $PA=\frac{1}{2}PO$

$2PA=PO2k=\sqrt{h^2+k^2}4k^2=h^2+k^2{h}^2=3k^2$

Replacing $h$ by $x$ and $k$ by $y$

$x^2=3y^2$

is the required locus.