Question

Find the locus of a point equidistant from the point (2,4) and the y-axis.

Answer 1

$\text{Let P(h,k) be the point which is equidistant from the point (2,4) and the y-axis .The distance of point P(h,k) from the y-axis is h.}\therefore h=\sqrt{(h-2{)}^2+(k-4{)}^2}\Rightarrow h^2-4h-4+k^2-8k+16=h^2\Rightarrow k^2-4h-8k+20=0Hence,thelocusof(h,k)is{y}^2-4x-8y+20=0$