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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)
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