The correct option is C (x+1)2=4(y−1)
Any point on y2=4x is P(t2,2t).
Let the image of P about x−y+1=0 be (h,k).
⇒h=2t−1 and k=t2+1
Eliminating t, we get
Hence, locus of P is (x+1)2=4(y−1)