**
Question 18
**

Find the equations of the normal to the curve y = 4x
^{
3
}
– 3x + 5 which are perpendicular to the line 9x – y + 5 = 0

Last updated at Nov. 1, 2018 by Teachoo

**
Question 18
**

Find the equations of the normal to the curve y = 4x
^{
3
}
– 3x + 5 which are perpendicular to the line 9x – y + 5 = 0

Transcript

Question 18 Find the equations of the normal to the curve y = 4x3 – 3x + 5 which are perpendicular to the line 9x – y + 5 = 0 We know that Slope of normal × Slope of tangent = – 1 Slope of normal × 𝑑𝑦/𝑑𝑥 = – 1 Slope of normal = (−1)/(𝑑𝑦/𝑑𝑥) Finding 𝒅𝒚/𝒅𝒙 y = 4x3 – 3x + 5 Differentiating w.r.t x 𝑑𝑦/𝑑𝑥 = 12x2 – 3 Thus, Slope of normal = (−1)/(𝑑𝑦/𝑑𝑥) = (−1)/(12𝑥^2 − 3) Now, given that Normal is perpendicular to the line 9x – y + 5 = 0 Finding slope of 9x – y + 5 = 0 9x – y + 5 = 0 9x + 5 = y y = 9x + 5 Slope = 9 Since Normal is perpendicular to the line 9x – y + 5 = 0 Slope of normal × Slope of line = –1 (−1)/(12𝑥^2 − 3) × 9 = –1 1/(12𝑥^2 − 3) × 9 = 1 9 = 12x2 – 3 12x2 – 3 = 9 12x2 = 9 + 3 12x2 = 12 x2 = 12/12 x2 = 1 x = ± 1 Finding points for x = 1, –1 When x = 1 y = 4x3 – 3x + 5 y = 4(1)3 – 3(1) + 5 y = 4 – 3 + 5 y = 6 ∴ Point is (1, 6) When x = –1 y = 4x3 – 3x + 5 y = 4(–1)3 – 3(–1) + 5 y = –4 + 3 + 5 y = 4 ∴ Point is (–1, 4) Also, sine normal is perpendicular to line with slope 9 Slope of normal × 9 = –1 Slope of normal = (−1)/9 Finding equation of normals We know that Equation of line at (𝑥1 ,𝑦1) & having slope at 𝑚 is (𝑦−𝑦1)=𝑚(𝑥−𝑥1) Equation of normal, passing through (1, 6) with slope (−𝟏)/𝟗 (y – 6) = (−1)/9 (x – 1) 9(y – 6) = –1(x – 1) 9y – 54 = –x + 1 9y + x = 55 Equation of normal, passing through (–1, 4) with slope (−𝟏)/𝟗 (y – 4) = (−1)/9 (x – (–1)) 9(y – 4) = –1(x + 1) 9y – 36 = –x – 1 9y + x = 35

CBSE Class 12 Sample Paper for 2019 Boards

Paper Summary

Question 1

Question 2

Question 3

Question 4 (Or 1st)

Question 4 (Or 2nd)

Question 5

Question 6

Question 7

Question 8 (Or 1st)

Question 8 (Or 2nd)

Question 9

Question 10 (Or 1st)

Question 10 (Or 2nd)

Question 11

Question 12 (Or 1st)

Question 12 (Or 2nd)

Question 13 (Or 1st)

Question 13 (Or 2nd)

Question 14

Question 15

Question 16 (Or 1st)

Question 16 (Or 2nd)

Question 17

Question 18 You are here

Question 19

Question 20

Question 21 (Or 1st)

Question 21 (Or 2nd)

Question 22

Question 23

Question 24 (Or 1st)

Question 24 (Or 2nd)

Question 25

Question 26 (Or 1st)

Question 26 (Or 2nd)

Question 27 (Or 1st)

Question 27 (Or 2nd)

Question 28

Question 29

Class 12

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.