Fint the shortest distance of the point (0,c) from the parobola y=x² where 0≤c≤5. ANSWER-½√(4c-1)

Let there be a point on the parabola (x,y) which is closest to the given point.

Then the distance this point from (x,y)= √(x-0)² + (y-c)² .Substituting the value of x² with y the  equation becomes √y + (y-c)²=√y + y²+c²-2yc Differentiating the equ with respect to y

We get (1+2y-2c)/(2(√y + y²+c²-2yc))=0 ==> y=(2c-1)/2=c-(1/2)

Then x=√c-(1/2). The distance of the point =√(c-(1/2) + (c-(1/2)-c)²)=√c-(1/2) + (1/4) =√c-(1/4)=½√(4c-1)