Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find the equation of the normals to the curve y = x ^ 3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0.

21) Find the equation of the normals to the curvey = x^3 + 2x + 6 which are parallel
to the line x + 14y + 4 = 0.

Answers (1)

best_answer

Equation of given curve is
y = x^3 + 2x + 6
Parellel to line x + 14y + 4 = 0 \Rightarrow y = \frac{-x}{14} -\frac{4}{14} means slope of normal and line is equal
We know that, equation of line
y= mx + c
on comparing it with  our given equation. we get,
m = \frac{-1}{14}
Slope of tangent = \frac{dy}{dx} = 3x^2+2
We know that
Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent} = \frac{-1}{3x^2+2}
\frac{-1}{3x^2+2} = \frac{-1}{14}
3x^2+2 = 14\\ 3x^2 = 12 \\ x^2 = 4\\ x = \pm 2
Now, when x = 2,  y = (2)^3 + 2(2) + 6 = 8+4+6 =18
and 
When x = -2 , y = (-2)^3 + 2(-2) + 6 = -8-4+6 =-6
Hence, the coordinates are (2,18) and (-2,-6)
Now, the equation of at point (2,18) with slope \frac{-1}{14}
y-y_1=m(x-x_1)\\ y-18=\frac{-1}{14}(x-2)\\ 14y - 252 = -x + 2\\ x+14y = 254
Similarly,  the equation of at point (-2,-6) with slope \frac{-1}{14}

y-y_1=m(x-x_1)\\ y-(-6)=\frac{-1}{14}(x-(-2))\\ 14y + 84 = -x - 2\\ x+14y + 86= 0
Hence, the equation of the normals to the curvey = x^3 + 2x + 6 which are parallel
to the line x + 14y + 4 = 0. 

are x +14y - 254 = 0  and  x + 14y +86 = 0

Posted by

Gautam harsolia

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Board Exams 2025: Guidelines issued for Class 10, 12 practicals
anam-khan

Board Exam Date Sheet 2025 Live: ICSE, ISC, CBSE, MP time table out; BSEB routine soon
anam-khan

Board exam dates 2025: CBSE, CISCE, UP, MP, Maharashtra, TN time table out; state-wise updates

Latest Question