Question

Find the equation to the locus of a point which moves so that its distance from the point $(a,0)$ is always four times its distance from the axis of $y$.

Answer 1

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

Given point be $A(a,0)$

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

Distance of $P$ from $y$ axis be $PB$

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

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

Given $PA=4PB$

$\sqrt{{(h-a)}^2+{(k-0)}^2}=4h{(h-a)}^2+{\left(k\right)}^2=16h^2{h}^2+a^2-2ah+k^2=16h^{2}15h^2-k^2-a^2+2ah=0$

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

$15x^2-y^2-a^2+2ax=0{15x^2-y^2+2ax=a}^2$

is the required equation of locus