Application of Derivatives

Find the point on the curve x 2 = 4y which is nearest to the point (–1, 2). - Wired Faculty

Question

Find the point on the curve x² = 4y which is nearest to the point (–1, 2).

Solution

Let (x, y) be the point on $straight x squared space equals space 4 straight y$ which is nearest to the point (–1, 2).
$because space space space left parenthesis straight x comma space space straight y right parenthesis space lies space on space the space curve space straight y space equals space straight x squared over 4 space space space rightwards double arrow space space point space is space space open parentheses straight x comma space straight x squared over 4 close parentheses Let space straight d space be space the space distance space between space left parenthesis negative 1 comma space space 2 right parenthesis space and space open parentheses straight x comma space straight x squared over 4 close parentheses therefore space space space straight d space equals space square root of left parenthesis straight x plus 1 right parenthesis squared space plus space open parentheses straight x squared over 4 minus 2 close parentheses squared end root space space space rightwards double arrow space space space space space straight d squared space equals space left parenthesis straight x plus 1 right parenthesis squared plus open parentheses straight x squared over 4 minus 2 close parentheses squared Put space straight d squared space equals space straight D comma space space space space space space therefore space space space space straight D space equals space left parenthesis straight x plus 1 right parenthesis squared plus open parentheses straight x squared over 4 minus 2 close parentheses squared Now space straight d space is space maximum space or space minimumn space when space straight D space is space maximum space or space minimum.$
$dD over dx space equals space 2 left parenthesis straight x plus 1 right parenthesis plus 2 open parentheses straight x squared over 4 minus 2 close parentheses. space fraction numerator 2 straight x over denominator 4 end fraction space equals space 2 straight x plus 2 plus straight x cubed over 4 minus 2 straight x space equals space 2 plus straight x cubed over 4$
$dD over dx space equals space 0 space space give space us space 2 plus straight x cubed over 4 space equals space 0 space space space space or space space space 8 plus straight x cubed space equals space 0$
$therefore space space space space space space straight x cubed space equals space minus 8 space space space space space or space space space space straight x space equals space minus 2 space space space space space fraction numerator straight d squared straight D over denominator dx squared end fraction space space equals fraction numerator 3 straight x squared over denominator 4 end fraction$
At $straight x equals 2 comma space space space fraction numerator straight d squared straight D over denominator dx squared end fraction space equals space 3 over 4 cross times 4 space equals space 3 greater than 0$
$therefore space space space space space straight D space is space minimum space when space straight x space equals space 2 comma space space space straight y space equals 4 over 4 space equals space 1 therefore space space space space space required space point space is space left parenthesis negative 2 comma space 1 right parenthesis.$

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