The equation of given is y = e2x = x1 ... (i) putting x = 0 in eq (i) we get y = 1 ∴ The given point is P (0,1) on differentiating eq (i) wrt we get
dy
dx
= 2e2x+2x ∴ (
dy
dx
)(0,1) = 2 Equation of tangent at P (0 , 1) to eq (1) is y - 1 = 2 (x - 0) ⇒ 2x - y + 1 = 0 (ii) ∴ Required distance = length of perpendicular from (1,1) to the line 2x - y + 1 = 0 = 2−1+1√4+1 =