Question 27 (MCQ) - Tangents and Normals (using Differentiation) - Chapter 6 Class 12 Application of Derivatives

Last updated at April 16, 2024 by Teachoo

Ex 6.3, 27 - Line y = x + 1 is a tangent to y2 = 4x at - Ex 6.3

Ex 6.3,27 - Chapter 6 Class 12 Application of Derivatives - Part 2

Ex 6.3,27 - Chapter 6 Class 12 Application of Derivatives - Part 3

Transcript

Question 27 The line 𝑦=𝑥+1 is a tangent to the curve 𝑦2=4𝑥 at the point (A) (1, 2) (B) (2, 1) (C) (1, – 2) (D) (– 1, 2)Given Curve is 𝑦^2=4𝑥 Differentiating w.r.t. 𝑥 𝑑(𝑦^2 )/𝑑𝑥=𝑑(4𝑥)/𝑑𝑥 𝑑(𝑦^2 )/𝑑𝑦 × 𝑑𝑦/𝑑𝑥=4 2𝑦 × 𝑑𝑦/𝑑𝑥=4 𝑑𝑦/𝑑𝑥=4/2𝑦 𝑑𝑦/𝑑𝑥=2/𝑦 Given line is 𝑦=𝑥+1 The Above line is of the form 𝑦=𝑚𝑥+𝑐 when m is slope of line Slope of line 𝑦=𝑥+1 is 1 Now Slope of tangent = Slope of line 𝑑𝑦/𝑑𝑥=1 2/𝑦=1 2=𝑦 𝑦=2 Finding x when 𝑦=2 𝑦^2=4𝑥 (2)^2=4𝑥 4=4𝑥 4/4=𝑥 𝑥=1 Hence the Required point is (x, y) = (1 , 2) Correct Answer is (A)

Next: Approximations (using Differentiation) → Ask a doubt

Made by

Davneet Singh

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.

Class 6

Class 7

Class 8

Class 9

Class 10

Class 11

Class 12

Courses

Interesting

Popular