CBSE Class 12 Sample Paper for 2019 Boards

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

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

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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

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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.