Given curve is y=√x....(i) since, the point (a,b) is on the curve (i). Then, b=√a⇒b2=a....(ii) Let D= distance between (a,b) and (1,0) .
∴‌‌D2=(a−1)2+(b−0)2=a2+1−2a+b2
⇒‌‌D2=a2+1−2a+a  [from Eq. (ii)] ⇒‌‌D2=a2−a+1......(iii) Let P=a2−a+1 Now, P is max\/min according as D2 is max\/min ∴‌‌‌
dP
da
=2a−1 and ‌
d2p
da2
=2>0  (min) For maximum or minimum of P, put ‌
dP
da
=0⇒2a−1=0 ⇒‌‌a=‌
1
2
∴At(a=‌
1
2
), point (a,b) is closest to the point (1,0) From Eq. (ii). b2=‌