Application of Derivatives

Find the point on the curve x 2 = 8y which is nearest to the point (2, 4). - Wired Faculty

Question

Find the point on the curve x² = 8y which is nearest to the point (2, 4).

Solution

Let $left parenthesis straight x comma space straight y right parenthesis$ be the point on x² = 8y which is nearest to the point (2, 4).
$therefore space space space left parenthesis straight x comma space space straight y right parenthesis space lies space on space the space curve space space equals space straight x squared over 8 space space space space rightwards double arrow space space space point space is space open parentheses straight x comma space straight x squared over 8 close parentheses$
$Let space space straight d space be space the space distance space between space left parenthesis 2 comma space 4 right parenthesis space and space open parentheses straight x comma space space space straight x squared over 8 close parentheses therefore space space space space space straight d space equals space square root of left parenthesis straight x minus 2 right parenthesis squared plus open parentheses straight x squared over 8 minus 4 close parentheses squared end root space space space space rightwards double arrow space space space space straight d squared space equals space left parenthesis straight x minus 2 right parenthesis squared plus open parentheses straight x squared over 8 minus 4 close parentheses squared Put space straight d squared space equals space straight D comma space space space space space space space space therefore space space straight D space equals space left parenthesis straight x minus 2 right parenthesis squared plus open parentheses straight x squared over 8 minus 4 close parentheses squared$
Now d is maximum or minimum when D is maximum or minimum. $dD over dx space equals space 2 left parenthesis straight x minus 2 right parenthesis space plus space 2 space open parentheses straight x squared over 8 minus 4 close parentheses. space space fraction numerator 2 straight x over denominator 8 end fraction space equals space 2 straight x minus 4 plus straight x cubed over 16 minus 2 straight x space equals space minus 4 plus straight x cubed over 16$
$dD over dx space equals space 0 space space space space give space us space minus 4 plus straight x cubed over 16 space equals 0 space space space space space space or space space space straight x cubed space equals space 64 space space space space space rightwards double arrow space space space straight x space equals space 4 space space fraction numerator straight d squared straight D over denominator dx squared end fraction space equals space fraction numerator 3 straight x squared over denominator 16 end fraction At space straight x space equals space 4 comma space space fraction numerator straight d squared straight D over denominator dx squared end fraction space equals space fraction numerator 3 cross times 16 over denominator 16 end fraction space equals space 3 space greater than 0 therefore space space space space straight D space is space minimum space when space straight x space equals space 4 comma space space space straight y equals 16 over 8 space equals space 2$
$therefore space space space point space is space left parenthesis 4 comma space 2 right parenthesis$

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