Check sibling questions

Ex 6.3

Ex 6.3, 1 Deleted for CBSE Board 2023 Exams

Ex 6.3,2 Deleted for CBSE Board 2023 Exams

Ex 6.3,3 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,4 Deleted for CBSE Board 2023 Exams

Ex 6.3, 5 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,6 Deleted for CBSE Board 2023 Exams

Ex 6.3,7 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,8 Deleted for CBSE Board 2023 Exams

Ex 6.3,9 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,10 Deleted for CBSE Board 2023 Exams

Ex 6.3,11 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,12 Deleted for CBSE Board 2023 Exams

Ex 6.3,13 Deleted for CBSE Board 2023 Exams You are here

Ex 6.3, 14 (i) Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (ii) Important Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (iii) Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (iv) Important Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (v) Deleted for CBSE Board 2023 Exams

Ex 6.3,15 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,16 Deleted for CBSE Board 2023 Exams

Ex 6.3,17 Deleted for CBSE Board 2023 Exams

Ex 6.3,18 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,19 Deleted for CBSE Board 2023 Exams

Ex 6.3,20 Deleted for CBSE Board 2023 Exams

Ex 6.3,21 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,22 Deleted for CBSE Board 2023 Exams

Ex 6.3,23 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,24 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,25 Deleted for CBSE Board 2023 Exams

Ex 6.3,26 (MCQ) Important Deleted for CBSE Board 2023 Exams

Ex 6.3,27 (MCQ) Deleted for CBSE Board 2023 Exams

Ex 6.3, 13 - Find points on x2/9 + y2/16 = 1 at which tangents

Ex 6.3,13 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,13 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.3,13 - Chapter 6 Class 12 Application of Derivatives - Part 4 Ex 6.3,13 - Chapter 6 Class 12 Application of Derivatives - Part 5 Ex 6.3,13 - Chapter 6 Class 12 Application of Derivatives - Part 6 Ex 6.3,13 - Chapter 6 Class 12 Application of Derivatives - Part 7


Transcript

Ex 6.3, 13 Find points on the curve π‘₯^2/9 + 𝑦^2/16 = 1 at which the tangents are (i) parallel to x-axis (ii) parallel to y-axis. π‘₯^2/9 + 𝑦^2/16 = 1 𝑦^2/16=1βˆ’π‘₯^2/9 Differentiating w.r.t. π‘₯ 𝑑(𝑦^2/16)/𝑑π‘₯=𝑑(1βˆ’ π‘₯^2/9)/𝑑π‘₯ 1/16 𝑑(𝑦^2 )/𝑑π‘₯=𝑑(1)/𝑑π‘₯βˆ’π‘‘(π‘₯^2/9)/𝑑π‘₯ 1/16 Γ— 𝑑(𝑦^2 )/𝑑π‘₯ Γ— 𝑑𝑦/𝑑𝑦=0βˆ’1/9 𝑑(π‘₯^2 )/𝑑π‘₯ 1/16 Γ— 𝑑(𝑦^2 )/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯=(βˆ’ 1)/9 𝑑(π‘₯^2 )/𝑑π‘₯ 1/16 Γ—2𝑦 ×𝑑𝑦/𝑑π‘₯=(βˆ’ 1)/( 9) 2π‘₯ 𝑑𝑦/𝑑π‘₯=((βˆ’ 1)/( 9) 2π‘₯)/(1/16 2𝑦) 𝑑𝑦/𝑑π‘₯=(βˆ’ 16)/9 π‘₯/𝑦 Hence 𝑑𝑦/𝑑π‘₯=(βˆ’ 16)/9 π‘₯/𝑦 parallel to π‘₯βˆ’π‘Žπ‘₯𝑖𝑠 Given tangent is parallel to π‘₯βˆ’π‘Žπ‘₯𝑖𝑠 β‡’ Slope of tangent = Slope of π‘₯βˆ’π‘Žπ‘₯𝑖𝑠 𝑑𝑦/𝑑π‘₯=0 (βˆ’ 16)/( 9) π‘₯/𝑦=0 This is only possible if π‘₯=0 when π‘₯=0 π‘₯^2/4 + 𝑦^2/16=1 0/4+𝑦^2/16=1 (𝐼𝑓 𝑦=0, (βˆ’16)/( 9) π‘₯/0=∞) 𝑦^2/16=1 𝑦^2=16 𝑦=Β±4 Hence the points are (𝟎 , πŸ’) & (𝟎 , βˆ’πŸ’) parallel to π‘¦βˆ’π‘Žπ‘₯𝑖𝑠 Similarly if line is parallel to π‘¦βˆ’π‘Žπ‘₯𝑖𝑠 Angle with π‘₯βˆ’π‘Žπ‘₯𝑖𝑠 =90Β° ΞΈ = 90Β° Slope = tan ΞΈ = tan 90Β°=∞ Hence 𝑑𝑦/𝑑π‘₯=∞ 16/9 π‘₯/𝑦=∞ 16π‘₯/9𝑦=1/0 This will be possible only if Denominator is 0 9𝑦=0 𝑦=0 Now it is given that π‘₯^2/9+𝑦^2/16=1 Putting 𝑦=0 π‘₯^2/9+0/16=1 π‘₯^2/9=1 π‘₯^2=9 π‘₯=√9 π‘₯=Β±3 Hence the points at which tangent is parallel to π‘¦βˆ’π‘Žπ‘₯𝑖𝑠 are (πŸ‘ , 𝟎) & (βˆ’πŸ‘ , 𝟎)

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.