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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 March 29, 2023 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, Β± 4)Β  Β  Β  (b) (Β±4, 0)
(c) (Β±3, 0)Β  Β  Β  Β (d) (0, Β±3)

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

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Question 16 The points on the curve π‘₯^2/9+𝑦^2/16 = 1 at which the tangents are parallel to y-axis are: (a) (0, Β± 4) (b) (Β±4, 0) (c) (Β±3, 0) ` (d) (0, Β±3) π‘₯^2/9 + 𝑦^2/16 = 1 π’š^𝟐/πŸπŸ”=πŸβˆ’π’™^𝟐/πŸ— 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) π’…π’š/𝒅𝒙=(βˆ’ πŸπŸ”)/πŸ— 𝒙/π’š Since tangents parallel to y-axis ∴ Angle with x-axis = 90Β° ΞΈ = 90Β° Slope = tan ΞΈ = tan 90Β° = ∞ Hence π’…π’š/𝒅𝒙=∞ 16/9 π‘₯/𝑦=∞ πŸπŸ”π’™/πŸ—π’š=𝟏/𝟎 This will be possible only if Denominator is 0 9𝑦=0 π’š=𝟎 Finding value of x by putting y = 0 in equation 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 𝒙=Β±πŸ‘ Hence, Required points = (Β± 3, 0) So, the correct answer is (c)

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