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

dear student

as it is given that the axes of coordinates being drawn through the fixed point

so fixed point is origin (0,0)

coordinate axes are being parallel and perpendicular to the given line

A fixed point is at a perpendicular distance "a" from a fixed straight line
so if fixed point is (0,0)
then the straight line passes through either (a,0) or (-a,0)
and its equation is

$$\begin{gathered} x+a=0orx-a=0 \\ letthelocusofthemovingpointbe(h,k) \\ dis\tan ceofthispointfromfixedpoint(0,0)=dis\tan ceofthispointfromlinex+a=0orx-a=0 \\ so \\ \sqrt{h^2+k^2}=\frac{\left|h\pm a\right|}{\sqrt{1^2+0^2}} \\ squaringweget \\ {h}^2+k^2=h^2+a^2\pm 2ah \\ {k}^2=a^2\pm 2ah \\ so,locusis \\ {y}^2=a^2\pm 2ax \end{gathered}$$



regards