Q. 185.0( 4 Votes )

Find the eq

Answer :

Curve: y = 4x3 – 3x +5


Let the required normal be at (x1, y1) on the curve


Slope of tangent of the curve




Slope of normal at any point (x1, y1)


Slope of given line = 9


As, given line is perpendicular to the normal, therefore,






Now, put x1 = +1 and -1 on the equation of curve to find y1


Equation of curve: y = 4x3 – 3x +5


Put x = 1, we get,


y = 4(1)3 – 3(1) + 5


y = 4 – 3 + 5


y = 6


So, point is (1, 6)


Now put x = -1 in equation of curve we get,


y = 4(-1)3 – 3(-1) + 5


y = -4 + 3 + 5


y = 4


So, point is (-1, 4)


So, we get the points as (1, 6) and (-1, 4)


Therefore, equations of normal are:



9y – 54 = –x + 1 and 9y – 36 = –x – 1


x + 9y = 55 and x + 9y = 35


So, required equations are x + 9y = 55 and x + 9y = 35.


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