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Application Of Derivatives

Question
CBSEENMA12035201
Wired Faculty App

Find the equation of the normals to the curve y = x3 + 2x + 6 which are parallel to the line 14y + 4 = 0.

Solution

The equation of curve is  y = x3 + 2x + 6                           ...(1)
                                  dy over dx space equals space 3 straight x squared plus 2
therefore space space space space at space left parenthesis straight x comma space straight y right parenthesis space space slope space of space tangent space space equals space 3 straight x squared plus 2 therefore space space space space space at space left parenthesis straight x comma space straight y right parenthesis comma space slope space of space normal space equals space minus fraction numerator 1 over denominator 3 straight x squared plus 2 end fraction therefore space space space space space space space straight m subscript 1 space equals space minus fraction numerator 1 over denominator 3 straight x squared plus 2 end fraction
Let straight m subscript 2 be slope of line straight x plus 14 straight y plus 4 space equals space 0                                    ...(2)
therefore space space space space space straight m subscript 2 space equals space minus 1 over 14
because space space space space normal space is space parallel space to space left parenthesis 2 right parenthesis comma space space space space space space space therefore space space space space straight m subscript 1 space equals space straight m subscript 2
therefore space space space space space space space space minus fraction numerator 1 over denominator 3 straight x squared plus 2 end fraction space equals space minus 1 over 14 space space space space space rightwards double arrow space space space space space 3 straight x squared plus 2 space equals space 14 space space space rightwards double arrow space space space space 3 straight x squared space space equals space 12 rightwards double arrow 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 straight x squared space equals space 4 space space space space space space space space space space space space space space space space space rightwards double arrow space space space space space straight x space equals space minus 2 comma space 2 therefore space space space space space space space space space space space space space from space left parenthesis 1 right parenthesis comma space space space straight y space equals space left parenthesis negative 2 right parenthesis cubed plus 2 left parenthesis negative 2 right parenthesis space plus space 6 comma space space space left parenthesis 2 right parenthesis cubed plus 2 left parenthesis 2 right parenthesis space plus space 6 space equals space minus 6 comma space space 18 therefore space space space space space points space are space left parenthesis negative 2 comma space minus 6 right parenthesis comma space left parenthesis 2 comma space 18 right parenthesis The space equation space of space normal space at space left parenthesis negative 2 comma space space minus 6 right parenthesis space is space space space space space space space space space space space space space space space space straight y plus 6 space equals space minus 1 over 14 left parenthesis straight x plus 2 right parenthesis comma space space space space or space space space 14 straight y plus 84 space equals negative straight x minus 2 space space space space or space space space space space straight x plus 14 straight y plus 86 space equals space 0 The space equation space of space normal space at space left parenthesis 2 comma space 18 right parenthesis thin space is space space space straight y minus 18 space equals space minus 1 over 14 left parenthesis straight x minus 2 right parenthesis space space space or space space 14 straight y minus 252 space equals space minus straight x plus 2 space space space space or space space space straight x plus 14 straight y minus 254 space equals space 0

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