Tangent and Normal Question6





In this page tangent and normal question6 we are going to see solution of some practice questions from the worksheet.

(6) Find the equations of normal to y = x³ - 3 x that is parallel to 2 x + 18 y - 9 = 0.

Solution:

Here the normal line drawn to the curve is parallel to the line 2 x + 18 y - 9 = 0

Slope of the line parallel to the normal line to the curve

             m = - coefficient of x/coefficient of y

             m = -2/18

                = -1/9   ----- (1)

 Slope of the line parallel to the normal line to the curve = -1/9

 Slope of the tangent to the curve y = x³ - 3 x

  differentiate with respect to "x"

                         dy/dx = 3 x² - 3 (1)

                                  = 3 x² - 3

slope of the normal line to the curve  = -1/(3 x² - 3) ----- (2)

       (1) = (2)

     -1/9 = -1/(3 x² - 3)

     3 x² - 3 = 9

        3 x² = 9 + 3

         3 x² = 12

           x² = 12/3

           x² = 4

           x = √4

           x = ± 2

to find the y-coordinate we have to apply the x values in the equation of the curve not in the given line 

 x = 2                                           x = -2

y = x³ - 3 x                                   y = x³ - 3 x

   = 2³ - 3 (2)                                   = (-2)³ - 3 (-2)

   = 8 - 6                                         = - 8 + 6

   = 2                                              = -2

point of contact are (2,2) (-2,-2)

Equation of the normal passing through the point (2,2) and slope is -1/9

 (y - y₁) = (-1/m) (x - x₁)

 (y - 2) = (-1/9) (x - 2)

9 (y - 2) = -1(x - 2)

9 y - 18 = - x + 2

 x + 9 y - 18 - 2 = 0

 x + 9 y - 20 = 0

Equation of the normal passing through the point (-2,-2) and slope is -1/9

 (y - y₁) = (-1/m) (x - x₁)

[y - (-2)] = (-1/9) [x - (-2)]

[y + 2] = (-1/9) [x + 2]

9 (y + 2) = -1 (x + 2)

9 y + 18 = - x - 2

 x + 9 y + 18 + 2 = 0

 x + 9 y + 20 = 0

tangent and normal question6 tangent and normal question6