**
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

CBSE Class 12 Sample Paper for 2019 Boards

Paper Summary

Question 1 Important

Question 2

Question 3

Question 4 (Or 1st) Important

Question 4 (Or 2nd)

Question 5

Question 6

Question 7 Important

Question 8 (Or 1st) Important

Question 8 (Or 2nd)

Question 9

Question 10 (Or 1st) Important

Question 10 (Or 2nd)

Question 11 Important

Question 12 (Or 1st)

Question 12 (Or 2nd)

Question 13 (Or 1st) Important

Question 13 (Or 2nd)

Question 14 Important

Question 15

Question 16 (Or 1st)

Question 16 (Or 2nd) Important

Question 17

Question 18 You are here

Question 19 Important

Question 20 Important

Question 21 (Or 1st)

Question 21 (Or 2nd) Important

Question 22

Question 23 Important

Question 24 (Or 1st)

Question 24 (Or 2nd) Important

Question 25

Question 26 (Or 1st) Important

Question 26 (Or 2nd)

Question 27 (Or 1st) Important

Question 27 (Or 2nd) Important

Question 28

Question 29 Important

Class 12

Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Oct. 1, 2019 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

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