Let the image of `(lambda^(2), 2 lambda)` in the line mirror x-y+1=0 be (h,k).
`therefore (h-lambda^(2))/(1) = (k-2lambda)/(-1) = (-2(lambda^(2)-2lambda +1))/(2)`
`therefore h+1 = 2lambda " " (1)`
`" and " k= lambda^(2) + 1 " " (2)`
Putting the value of `lambda` from (1) in (2), we get
`k-1 = ((h+1)/(2))^(2)`
or 4(y-1) = `(x+1)^(2)`
This is the required locus.