find the equation of normal to the curve Y is equals to x cube + x square - 11 x + 15 where normal is parallel to the line x - 3 Y + 1 equals to zero
Answers
we have to find the equation of normal to the curve y = x³ + x² - 11x + 15 where normal is parallel to the line x - 3y + 1 = 0.
solution : curve y = x³ + x² - 11x + 15
slope of tangent of curve = dy/dx = 3x² + 2x - 11
we know, slope of normal = -1/slope of tangent
= -1/(3x² + 2x - 11)
but normal to the curve is parallel to the line x - 3y + 1 = 0
so, slope of normal = slope of line
⇒-1/(3x² + 2x - 11) = 1/3
⇒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 , -2
at x = 4/3, y = (4/3)³ + (4/3)² - 11 × 4/3 + 15 = 121/27
at x = -2, y = (-2)³ + (-2)² - 11(-2) + 15 = -8 + 4 + 22 + 15 = 33
so points are (4/3, 121/27) and (-2, 33)
equation of normal to the curve :
(y - 33) = 1/3(x + 2)
⇒x + 2 - 3y + 99 = 0
⇒x - 3y + 101 = 0
again, (y - 121/27) = 1/3(x - 4/3)
⇒3y - 121/9 = x - 4/3
⇒x - 3y - 4/3 + 121/9 = 0
⇒x - 3y + 109/9 = 0
Answer :-
we have to find the equation of normal to the curve y = x³ + x² - 11x + 15 where normal is parallel to the line x - 3y + 1 = 0.
solution : curve y = x³ + x² - 11x + 15
slope of tangent of curve = dy/dx = 3x² + 2x - 11
we know, slope of normal = -1/slope of tangent
= -1/(3x² + 2x - 11)
but normal to the curve is parallel to the line x - 3y + 1 = 0
so, slope of normal = slope of line
⇒-1/(3x² + 2x - 11) = 1/3
⇒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 , -2
at x = 4/3, y = (4/3)³ + (4/3)² - 11 × 4/3 + 15 = 121/27
at x = -2, y = (-2)³ + (-2)² - 11(-2) + 15 = -8 + 4 + 22 + 15 = 33
so points are (4/3, 121/27) and (-2, 33)
equation of normal to the curve :
(y - 33) = 1/3(x + 2)
⇒x + 2 - 3y + 99 = 0
⇒x - 3y + 101 = 0
again, (y - 121/27) = 1/3(x - 4/3)
⇒3y - 121/9 = x - 4/3
⇒x - 3y - 4/3 + 121/9 = 0
⇒x - 3y + 109/9 = 0