Minimum distance between the curves y 2 = x - 1 and x 2 = y - 1 is equal to

Step-by-step solution

General point on the curve y 2 = x - 1 is ( t 1 2 + 1 , t 1 ) and the general point on the curve x 2 = y - 1 is ( t 2 , t 2 2 + 1 ) . Since both the curves are symmetrical about the line y = x , for the nearest point on the curve y 2 = x - 1 from the line y = x Let, Q t 1 2 + 1, t 1 P ( t 2 , t 2 2 + 1 ) slopes of the tangents are same 1 2 t 1 = 2 ⋅ t 2 t 1 t 2 = 1 4 ......(1) Since P Q is perpendicular to line x = y t 1 - t 2 2 - 1 t 1 2 + 1 - t 2 = - 1 ......(2) Solving (1) & (2), we get t 1 = t 2 = 1 2 P 1 2 , 5