Application of Derivatives

Find the shortest distance of the point (0, c) from the parabola y = x 2 , where 0 ≤ c ≤ 5. - Wired Faculty

Question

Find the shortest distance of the point (0, c) from the parabola y = x² , where 0 ≤ c ≤ 5.

Solution

Let x be the distance between (0, c) and any point (x, y) on the parabola y =x^2. $therefore space space space space space straight s space equals space square root of left parenthesis straight x minus 0 right parenthesis squared plus left parenthesis straight y minus straight c right parenthesis squared end root space equals space square root of straight x squared plus left parenthesis straight y minus straight c right parenthesis squared end root rightwards double arrow space space space straight s space equals space square root of straight y plus left parenthesis straight y minus straight c right parenthesis squared end root space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space straight y space equals straight x squared close square brackets rightwards double arrow space space space space ds over dy space equals space fraction numerator 1 over denominator 2 square root of straight y plus left parenthesis straight y minus straight c right parenthesis squared end root end fraction left square bracket 1 plus 2 space left parenthesis straight y minus straight c right parenthesis right square bracket space equals space fraction numerator 2 straight y plus left parenthesis 1 minus 2 straight c right parenthesis over denominator 2 square root of straight y plus left parenthesis straight y minus straight c right parenthesis squared end root end fraction space space space space space space space ds over dy space equals space 0 space space space space space space space rightwards double arrow space space space space space 2 straight y left parenthesis 1 minus 2 straight c right parenthesis space equals space 0 space space space space space space space space space rightwards double arrow space straight y space equals space straight c minus 1 half space space space space space$
$When space straight y thin space less than space straight c minus 1 half space slightly comma space ds over dy space equals space minus ve and space when space straight y greater than straight c space minus 1 half space space slightly comma space space space space ds over dy space equals space plus ve therefore space space space space space space at space space straight y space equals space straight c minus 1 half comma space space ds over dy space changes space from space minus ve space to space ve therefore space space space space space straight s space is space minimum space at space straight y space equals space straight c minus 1 half.$

$and space minimum space value space of space straight s space equals space square root of open parentheses straight c minus 1 half close parentheses space plus space open parentheses straight c minus 1 half minus straight c close parentheses squared end root space equals space square root of straight c minus 1 half plus 1 fourth end root$
$equals square root of straight c minus 1 fourth end root space equals space square root of fraction numerator 4 straight c minus 1 over denominator 4 end fraction end root space equals fraction numerator space square root of 4 straight c minus 1 end root over denominator 2 end fraction.$

For Daily Free Study Material Join wiredfaculty