Application of Derivatives

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

Question

Find the point on the curve y² = 2x which is nearest to the point (1, – 4).

Solution

Any point on the parabola $straight y squared space equals 2 straight x$ is $straight P open parentheses 1 half straight t squared comma space space straight t close parentheses$ , Let $straight Q space be thin space left parenthesis 1 comma space minus 4 right parenthesis$
Let d be the distance between Q(1, -4) and $straight P open parentheses 1 half straight t squared comma space space straight t close parentheses$
$therefore space space space straight d space equals space square root of open parentheses straight t squared over 2 minus 1 close parentheses squared. space plus left parenthesis straight t plus 4 right parenthesis squared end root space space space space space space space space space space space space rightwards double arrow space space space space space space straight d squared space equals space open parentheses straight t squared over 2 minus 1 close parentheses squared space plus space left parenthesis straight t plus 4 right parenthesis squared therefore space space space space straight D space equals space straight t to the power of 4 over 4 minus straight t squared plus 1 plus straight t squared plus 8 straight t plus 16 space equals space straight t to the power of 4 over 4 plus 8 straight t plus 17 comma space space where space straight D space equals space straight d squared Now space straight d space is space maximum space or space minimum space when space straight D space is space maximum space or space minimum.$
$dD over dt space equals straight t cubed plus 8 space space dD over dt space equals space 0 space space space space space space rightwards double arrow space space space space straight t cubed space equals space minus 8 space space space space rightwards double arrow space space space straight t space equals space minus 2 space fraction numerator straight d squared straight D over denominator dt squared end fraction space equals space 3 straight t squared.$
$At space straight t space equals space minus 2 comma space space space space space fraction numerator straight d squared straight D over denominator dt squared end fraction space equals space 3 left parenthesis negative 2 right parenthesis squared space equals space 12 space greater than 0 space therefore space space space straight D space is space minimum space when space straight t space equals space minus 2 therefore space space space space straight P space is space left parenthesis 2 comma space minus 2 right parenthesis comma space which space is space nearest space to space straight Q left parenthesis 1 comma space minus 4 right parenthesis$

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