Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Finding equation of tangent/normal when point and curve is given
Question 14 (i) Deleted for CBSE Board 2024 Exams You are here
Question 22 Deleted for CBSE Board 2024 Exams
Question 24 Important Deleted for CBSE Board 2024 Exams
Question 7 Deleted for CBSE Board 2024 Exams
Question 5 Deleted for CBSE Board 2024 Exams
Question 13 Important Deleted for CBSE Board 2024 Exams
Question 2 Deleted for CBSE Board 2024 Exams
Question 6 (MCQ) Deleted for CBSE Board 2024 Exams
Question 7 (MCQ) Important Deleted for CBSE Board 2024 Exams
Question 20 Deleted for CBSE Board 2024 Exams
Question 8 (MCQ) Important Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Finding equation of tangent/normal when point and curve is given
Last updated at May 29, 2023 by Teachoo
Question 14 Find the equations of the tangent and normal to the given curves at the indicated points: (i) 𝑦=𝑥4 −6𝑥3+13𝑥2 −10𝑥+5 𝑎𝑡 (0, 5) 𝑦=𝑥4 −6𝑥3+13𝑥2 −10𝑥+5 Differentiating w.r.t. 𝑥 𝑑𝑦/𝑑𝑥=4𝑥^3−18𝑥^2+26𝑥−10 Now Point Given is (0 ,5) Hence 𝑥=0 , 𝑦=5 Putting 𝑥=0 in (1) Slope of tangent at (0 , 5) 〖𝑑𝑦/𝑑𝑥│〗_((0, 5) )=4(0)^3−18(0)^2+26(0)−10 〖𝑑𝑦/𝑑𝑥│〗_((0, 5) )=0−0+0−10 𝑑𝑦/𝑑𝑥=−10 Hence, Slope of tangent =−10 We know that Slope of tangent × Slope of Normal =−1 −10 ×"Slope of Normal "=−1 "Slope of Normal" =(−1)/(−10)=1/10 Hence Slope of tangent at (0, 5)=−10 & Slope of Normal at (0, 5)=1/10 Finding equation of tangent & normal Now Equation of line at (𝑥1 , 𝑦1) & having Slope m is 𝑦−𝑦1=𝑚(𝑥−𝑥1) Equation of tangent at (0, 5) & Slope –10 is (𝑦−5)=−10(𝑥−0) 𝑦−5=−10𝑥 10𝑥+𝑦−5=0 𝟏𝟎𝒙+𝒚=𝟓 Equation of Normal at (0, 5) & Slope 1/10 is (𝑦−5)=1/10 (𝑥−0) 𝑦−5=1/10 𝑥 10(𝑦−5)=𝑥 10𝑦−50=𝑥 𝒙−𝟏𝟎𝒚+𝟓𝟎=𝟎