Question

A fixed point is at a perpendicular distance a from a fixed straight line and a point moves so that its distance from the fixed point is always equal to its distance from the fixed line. Find the equation to its locus, the axes of coordinates being drawn through the fixed point and being parallel and perpendicular to the given line.

Answer 1

let the variable point be $P(h,k)$

The coordinate axes is drawn through the fixed point , so the fixed point is $O(0,0)$

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

The fixed line is parallel and perpendicular to the axes which tells us that it is parallel to $x$ axis and perpendicular to $y$ axis

It is at a distance $a$ from the fixed point

So its equation is $y=a$

Distance of $P$ from this line be $PA$

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

Given $PO=PA$

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

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

$x^2+2ay-a^2=0$

is the required locus.