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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Differential Calculus> Maxima and minima...

Fint the shortest distance of the point (0,c) from the parobola y=x²where 0≤c≤5. ANSWER-½√(4c-1)

shaleen upadhyay , 13 Years ago

Grade 12

1 Answers

Swapnil Saxena

Last Activity: 13 Years ago

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 x²with 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)