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

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.

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