We have y2 = 4x
Substituting the value of y = mx + 1 in y2 = 4x, we get
(mx + 1)2 = 4x
⇒ m2x2 + 2mx + 1 = 4x
⇒ m2x2 + (2m − 4)x + 1 = 0 .....(1)
Since, a tangent touches the curve at a point, the roots of (1) must be equal.
∴ D = 0
⇒ (2m − 4)2 − 4m2 = 0
⇒ 4m2 −16m + 16 − 4m2 = 0
⇒ m = 1