Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Tangents and Normals (using Differentiation)
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 4 Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Question 6 Deleted for CBSE Board 2024 Exams
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 8 Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Question 10 Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
Question 12 Deleted for CBSE Board 2024 Exams
Question 13 Deleted for CBSE Board 2024 Exams
Question 14 (i) Deleted for CBSE Board 2024 Exams
Question 14 (ii) Important Deleted for CBSE Board 2024 Exams
Question 14 (iii) Deleted for CBSE Board 2024 Exams
Question 14 (iv) Important Deleted for CBSE Board 2024 Exams
Question 14 (v) Deleted for CBSE Board 2024 Exams
Question 15 Important Deleted for CBSE Board 2024 Exams
Question 16 Deleted for CBSE Board 2024 Exams
Question 17 Deleted for CBSE Board 2024 Exams
Question 18 Important Deleted for CBSE Board 2024 Exams
Question 19 Deleted for CBSE Board 2024 Exams
Question 20 Deleted for CBSE Board 2024 Exams
Question 21 Important Deleted for CBSE Board 2024 Exams You are here
Question 22 Deleted for CBSE Board 2024 Exams
Question 23 Important Deleted for CBSE Board 2024 Exams
Question 24 Important Deleted for CBSE Board 2024 Exams
Question 25 Deleted for CBSE Board 2024 Exams
Question 26 (MCQ) Important Deleted for CBSE Board 2024 Exams
Question 27 (MCQ) Deleted for CBSE Board 2024 Exams
Tangents and Normals (using Differentiation)
Last updated at May 29, 2023 by Teachoo
Question 21 Find the equation of the normal to the curve 𝑦=𝑥^3+2𝑥+6 which are parallel to the line 𝑥+14𝑦+4=0.Let (ℎ , 𝑘) be the point on the Curve at which Normal is to be taken Given Curve is 𝑦=𝑥^3+2𝑥+6 Since point (ℎ , 𝑘) is on the Curve ∴ (ℎ , 𝑘) will satisfies the Equation of Curve Putting 𝑥=ℎ , 𝑦=𝑘 𝑘=ℎ^3+2ℎ+6 We know that Slope of a tangent to the Curve is 𝑑𝑦/𝑑𝑥 𝑦=𝑥^3+2𝑥+6 Differentiating w.r.t. 𝑥 𝑑𝑦/𝑑𝑥=3𝑥^2+2 Since tangent to be taken from (ℎ , 𝑘) Slope of tangent at (ℎ , 𝑘) is 〖𝑑𝑦/𝑑𝑥│〗_((ℎ, 𝑘) )=3ℎ^2+2 We know that Slope of tangent × Slope of Normal =−1 (3ℎ^2+2) × Slope of Normal =−1 Slope of Normal = (−1)/(3ℎ^2 + 2) Also, Given that Normal is parallel to the line 𝑥+14𝑦+4=0 If two lines are parallel then slopes are equal ⇒ Slopes of Normal = Slope of line 𝑥+14𝑦+4=0 Now, line is 𝑥+14𝑦+4=0 14𝑦=−𝑥−4 𝑦=(− 𝑥 − 4)/14 𝑦=((−1)/14)𝑥−(4/14) The above equation is of the form 𝑦=𝑚𝑥+𝑐 where m is slope ∴ Slope of line 𝑥+14𝑦+4=0 is (−1)/14 Now, Slope of Normal = Slope of line 𝑥+14𝑦+4=0 (−1)/(3ℎ^2 + 2)=(−1)/( 14) 1/(3ℎ^2 + 2)=1/( 14) 14=3ℎ^2+2 3ℎ^2+2=14 3ℎ^2=14−2 3ℎ^2=12 ℎ^2=12/3 ℎ^2=4 ℎ=±√4 ℎ=±2 When 𝒉=𝟐 𝑘=ℎ^3+2ℎ+6 𝑘=(2)^3+2(2)+6 𝑘=8+4+6 𝑘=18 ∴ Point is (𝟐 , 𝟏𝟖) When 𝒉=−𝟐 𝑘=ℎ^3+2ℎ+6 𝑘=(−2)^3+2(−2)+6 𝑘=−8−4+6 𝑘=− 6 ∴ Point is (−𝟐, −𝟔) Finding equation of normal We know that Equation of line at (𝑥1 , 𝑦1)& having Slope m is 𝑦−𝑦1=𝑚(𝑥−𝑥1) Equation of 1st Normal at (2 , 18) & having Slope (−1)/14 is (𝑦−18)=(−1)/14 (𝑥−2) 14(𝑦−18)=−(𝑥−2) 14𝑦−252=−𝑥+2 14𝑦+𝑥−252−2=0 𝒙+𝟏𝟒𝒚−𝟐𝟓𝟒=𝟎 Equation of 2nd Normal at (−2 , −6) & having Slope (−1)/14 is (𝑦−(−6))=(−1)/14 (𝑥−(−2)) 𝑦+6=(−1)/14 (𝑥+2) 14𝑦+84=−𝑥−2 14𝑦+𝑥+84+2=0 𝒙+𝟏𝟒𝒚+𝟖𝟔=𝟎