The points on the curve x 2 /9+y 2 /16 = 1 at which the tangents are parallel to y-axis are:
(a) (0, ± 4)      (b) (±4, 0)
(c) (±3, 0)       (d) (0, ±3)

This question is inspired from Ex 6.3,13 - Chapter 6 Class 12 - Application of Derivatives

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  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Transcript

Question 16 The points on the curve ๐‘ฅ^2/9+๐‘ฆ^2/16 = 1 at which the tangents are parallel to y-axis are: (a) (0, ยฑ 4) (b) (ยฑ4, 0) (c) (ยฑ3, 0) ` (d) (0, ยฑ3) ๐‘ฅ^2/9 + ๐‘ฆ^2/16 = 1 ๐’š^๐Ÿ/๐Ÿ๐Ÿ”=๐Ÿโˆ’๐’™^๐Ÿ/๐Ÿ— Differentiating w.r.t. ๐‘ฅ ๐‘‘(๐‘ฆ^2/16)/๐‘‘๐‘ฅ=๐‘‘(1โˆ’ ๐‘ฅ^2/9)/๐‘‘๐‘ฅ 1/16 ๐‘‘(๐‘ฆ^2 )/๐‘‘๐‘ฅ=๐‘‘(1)/๐‘‘๐‘ฅโˆ’๐‘‘(๐‘ฅ^2/9)/๐‘‘๐‘ฅ 1/16 ร— ๐‘‘(๐‘ฆ^2 )/๐‘‘๐‘ฅ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฆ=0โˆ’1/9 ๐‘‘(๐‘ฅ^2 )/๐‘‘๐‘ฅ 1/16 ร— ๐‘‘(๐‘ฆ^2 )/๐‘‘๐‘ฆ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=(โˆ’ 1)/9 ๐‘‘(๐‘ฅ^2 )/๐‘‘๐‘ฅ 1/16 ร— 2๐‘ฆ ร—๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=(โˆ’ 1)/( 9) 2๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=((โˆ’ 1)/( 9) 2๐‘ฅ)/(1/16 2๐‘ฆ) ๐’…๐’š/๐’…๐’™=(โˆ’ ๐Ÿ๐Ÿ”)/๐Ÿ— ๐’™/๐’š Since tangents parallel to y-axis โˆด Angle with x-axis = 90ยฐ ฮธ = 90ยฐ Slope = tan ฮธ = tan 90ยฐ = โˆž Hence ๐’…๐’š/๐’…๐’™=โˆž 16/9 ๐‘ฅ/๐‘ฆ=โˆž ๐Ÿ๐Ÿ”๐’™/๐Ÿ—๐’š=๐Ÿ/๐ŸŽ This will be possible only if Denominator is 0 9๐‘ฆ=0 ๐’š=๐ŸŽ Finding value of x by putting y = 0 in equation ๐‘ฅ^2/9+๐‘ฆ^2/16=1 Putting ๐‘ฆ=0 ๐‘ฅ^2/9+0/16=1 ๐‘ฅ^2/9=1 ๐‘ฅ^2=9 ๐‘ฅ=โˆš9 ๐’™=ยฑ๐Ÿ‘ Hence, Required points = (ยฑ 3, 0) So, the correct answer is (c)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.