find the equation of normal to the curve y=x^3+x^2-11+15 where normal is parallel to the line x- 3y+1=0.
Answers
Given : y = x³ + x² - 11x + 15
To Find : equation of normal to the curve parallel to the line x - 3y + 1 = 0.
Solution:
y = x³ + x² - 11x + 15
dy/dx = 3x² + 2x - 11
normal is parallel to the line x - 3y + 1 = 0.
3y = x + 1
=> y = x/3 + 1/3
Slope of normal = 1/3
Hence slope of Tangent = - 3
dy/dx = - 3
3x² + 2x - 11 = - 3
=> 3x² + 2x - 8 = 0
=> 3x² + 6x - 4x - 8 = 0
=> 3x(x + 2) - 4(x + 2) = 0
=> (3x - 4)(x + 2) = 0
=> x = 4/3 , x = -2
x = -2 => y = x³ + x² - 11x + 15 = 33
point ( -2, 33) slope = 1/3
y - 33 = (1/3)(x - (-2))
=> 3y - 99 = x + 2
=> x - 3y + 101 = 0
x = 4/3 => y = x³ + x² - 11x + 15 = 121/27
( 4/3 , 121/27)
y - 121/27 = (1/3)(x - 4/3)
=> 27y - 121 = 9 (x - 4/3)
=> 27y - 121 = 9x - 12
=> 9x - 27y + 109 = 0
x - 3y + 101 = 0
9x - 27y + 109 = 0
are Equations of Normal to the curve x³ + x² - 11x + 15
parallel to the line x - 3y + 1 = 0.
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