CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)

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Question 16 - CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at April 16, 2024 by

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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Transcript

Question 16 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) 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 Since tangents 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 Finding value of x by putting y = 0 in equation 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, Required points = (± 3, 0) So, the correct answer is (c)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.