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 You are here
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
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 13 Find points on the curve 𝑥^2/9 + 𝑦^2/16 = 1 at which the tangents are (i) parallel to x-axis (ii) parallel to y-axis. 𝑥^2/9 + 𝑦^2/16 = 1 𝑦^2/16=1−𝑥^2/9 Differentiating w.r.t. 𝑥 𝑑(𝑦^2/16)/𝑑𝑥=𝑑(1− 𝑥^2/9)/𝑑𝑥 1/16 𝑑(𝑦^2 )/𝑑𝑥=𝑑(1)/𝑑𝑥−𝑑(𝑥^2/9)/𝑑𝑥 1/16 × 𝑑(𝑦^2 )/𝑑𝑥 × 𝑑𝑦/𝑑𝑦=0−1/9 𝑑(𝑥^2 )/𝑑𝑥 1/16 × 𝑑(𝑦^2 )/𝑑𝑦 × 𝑑𝑦/𝑑𝑥=(− 1)/9 𝑑(𝑥^2 )/𝑑𝑥 1/16 ×2𝑦 ×𝑑𝑦/𝑑𝑥=(− 1)/( 9) 2𝑥 𝑑𝑦/𝑑𝑥=((− 1)/( 9) 2𝑥)/(1/16 2𝑦) 𝑑𝑦/𝑑𝑥=(− 16)/9 𝑥/𝑦 Hence 𝑑𝑦/𝑑𝑥=(− 16)/9 𝑥/𝑦 parallel to 𝑥−𝑎𝑥𝑖𝑠 Given tangent is parallel to 𝑥−𝑎𝑥𝑖𝑠 ⇒ Slope of tangent = Slope of 𝑥−𝑎𝑥𝑖𝑠 𝑑𝑦/𝑑𝑥=0 (− 16)/( 9) 𝑥/𝑦=0 This is only possible if 𝑥=0 when 𝑥=0 𝑥^2/4 + 𝑦^2/16=1 0/4+𝑦^2/16=1 (𝐼𝑓 𝑦=0, (−16)/( 9) 𝑥/0=∞) 𝑦^2/16=1 𝑦^2=16 𝑦=±4 Hence the points are (𝟎 , 𝟒) & (𝟎 , −𝟒) parallel to 𝑦−𝑎𝑥𝑖𝑠 Similarly if line is parallel to 𝑦−𝑎𝑥𝑖𝑠 Angle with 𝑥−𝑎𝑥𝑖𝑠 =90° θ = 90° Slope = tan θ = tan 90°=∞ Hence 𝑑𝑦/𝑑𝑥=∞ 16/9 𝑥/𝑦=∞ 16𝑥/9𝑦=1/0 This will be possible only if Denominator is 0 9𝑦=0 𝑦=0 Now it is given that 𝑥^2/9+𝑦^2/16=1 Putting 𝑦=0 𝑥^2/9+0/16=1 𝑥^2/9=1 𝑥^2=9 𝑥=√9 𝑥=±3 Hence the points at which tangent is parallel to 𝑦−𝑎𝑥𝑖𝑠 are (𝟑 , 𝟎) & (−𝟑 , 𝟎)