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

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

## 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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### Transcript

Question 16 The points on the curve π₯^2/9+π¦^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 + π¦^2/16 = 1 π^π/ππ=πβπ^π/π Differentiating w.r.t. π₯ π(π¦^2/16)/ππ₯=π(1β π₯^2/9)/ππ₯ 1/16 π(π¦^2 )/ππ₯=π(1)/ππ₯βπ(π₯^2/9)/ππ₯ 1/16 Γ π(π¦^2 )/ππ₯ Γ ππ¦/ππ¦=0β1/9 π(π₯^2 )/ππ₯ 1/16 Γ π(π¦^2 )/ππ¦ Γ ππ¦/ππ₯=(β 1)/9 π(π₯^2 )/ππ₯ 1/16 Γ 2π¦ Γππ¦/ππ₯=(β 1)/( 9) 2π₯ ππ¦/ππ₯=((β 1)/( 9) 2π₯)/(1/16 2π¦) ππ/ππ=(β ππ)/π π/π Since tangents parallel to y-axis β΄ Angle with x-axis = 90Β° ΞΈ = 90Β° Slope = tan ΞΈ = tan 90Β° = β Hence ππ/ππ=β 16/9 π₯/π¦=β πππ/ππ=π/π This will be possible only if Denominator is 0 9π¦=0 π=π Finding value of x by putting y = 0 in equation π₯^2/9+π¦^2/16=1 Putting π¦=0 π₯^2/9+0/16=1 π₯^2/9=1 π₯^2=9 π₯=β9 π=Β±π Hence, Required points = (Β± 3, 0) So, the correct answer is (c)