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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.