CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)
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CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)
Last updated at August 10, 2026 by Teachoo
This question is inspired from Ex 6.3,13 - Chapter 6 Class 12 - Application of Derivatives
Transcript
Question 16 The points on the curve š„^2/9+š¦^2/16 = 1 at which the tangents are parallel to y-axis are: (a) (0, Ā± 4) (b) (Ā±4, 0) (c) (Ā±3, 0) ` (d) (0, Ā±3) š„^2/9 + š¦^2/16 = 1 š^š/šš=šāš^š/š Differentiating w.r.t. š„ š(š¦^2/16)/šš„=š(1ā š„^2/9)/šš„ 1/16 š(š¦^2 )/šš„=š(1)/šš„āš(š„^2/9)/šš„ 1/16 Ć š(š¦^2 )/šš„ Ć šš¦/šš¦=0ā1/9 š(š„^2 )/šš„ 1/16 Ć š(š¦^2 )/šš¦ Ć šš¦/šš„=(ā 1)/9 š(š„^2 )/šš„ 1/16 Ć 2š¦ Ćšš¦/šš„=(ā 1)/( 9) 2š„ šš¦/šš„=((ā 1)/( 9) 2š„)/(1/16 2š¦) š š/š š=(ā šš)/š š/š Since tangents parallel to y-axis ā“ Angle with x-axis = 90Ā° Īø = 90Ā° Slope = tan Īø = tan 90Ā° = ā Hence š š/š š=ā 16/9 š„/š¦=ā ššš/šš=š/š This will be possible only if Denominator is 0 9š¦=0 š=š Finding value of x by putting y = 0 in equation š„^2/9+š¦^2/16=1 Putting š¦=0 š„^2/9+0/16=1 š„^2/9=1 š„^2=9 š„=ā9 š=Ā±š Hence, Required points = (Ā± 3, 0) So, the correct answer is (c)
