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The equation of the normal to the curve y = sin x at (0, 0) is:

(A)x = 0                  (B) y = 0

(C) x + y = 0           (D) x – y = 0


This question is similar to Ex 6.3, 14 (i) - Chapter 6 Class 12 - Application of Derivatives



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Question 4 The equation of the normal to the curve y = sin x at (0, 0) is: x = 0 (B) y = 0 (C) x + y = 0 (D) x – y = 0 𝑦=sin⁑π‘₯ Since Slope of normal =(βˆ’1)/(𝑑𝑦/𝑑π‘₯) Differentiating 𝑦 w.r.t. π‘₯ 𝑑𝑦/𝑑π‘₯=πœπ¨π¬β‘π’™ Since given point is (0, 0) Putting 𝒙=𝟎 in (1) π’…π’š/𝒅𝒙 =cos⁑0 𝑑𝑦/𝑑π‘₯=𝟏 Hence, Slope of normal =(βˆ’1)/(𝑑𝑦/𝑑π‘₯) =(βˆ’1)/1 =βˆ’πŸ Finding equation of normal Equation of line at (π‘₯1 , 𝑦1) & having Slope m is π‘¦βˆ’π‘¦1=π‘š(π‘₯βˆ’π‘₯1) ∴ Equation of Normal at (0, 0) & Slope -1 is (π’šβˆ’πŸŽ)=βˆ’πŸ(π’™βˆ’πŸŽ) 𝑦=βˆ’1(π‘₯) 𝑦=βˆ’π‘₯ π’š+𝒙=𝟎 Hence, equation of normal is π’š+𝒙=𝟎 So, the correct answer is (C)

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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.