Question 6 - Examples - Chapter 6 Class 12 Application of Derivatives

Last updated at August 2, 2026 by

Example 19 - Find equations of tangent, normal to x2/3 + y2/3 = 2

Example 19 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 19 - Chapter 6 Class 12 Application of Derivatives - Part 3 Example 19 - Chapter 6 Class 12 Application of Derivatives - Part 4

Take a fresh quiz. Then take another.
Every attempt is a new AI-adaptive Teachoo quiz with 2 questions, selected from your answers, mistakes, and progress.
Remove Ads

Transcript

Question 6 Find the equations of the tangent and normal to the curve š‘„^(2/3) + š‘¦^(2/3) = 2 at (1, 1).Given curve š‘„^(2/3) + š‘¦^(2/3) = 2 Differentiating both sides w.r.t x 2/3 š‘„^(1 āˆ’ 2/3)+2/3 š‘¦^(1 āˆ’ 2/3) š‘‘š‘¦/š‘‘š‘„ = 0 2/3 š‘„^((āˆ’1)/3)+2/3 š‘¦^((āˆ’1)/3) š‘‘š‘¦/š‘‘š‘„ = 0 2/3 š‘¦^((āˆ’1)/3) š‘‘š‘¦/š‘‘š‘„ = (āˆ’2)/3 š‘„^((āˆ’1)/3) 1/š‘¦^(1/3) š‘‘š‘¦/š‘‘š‘„ = (āˆ’1)/š‘„^(1/3) š’…š’š/š’…š’™ = āˆ’ (š’š/š’™)^(šŸ/šŸ‘) Thus, Slope of tangent to the curve = āˆ’ (š‘¦/š‘„)^(1/3) At point (1, 1) Slope = āˆ’ (šŸ/šŸ)^(šŸ/šŸ‘) = āˆ’1 Hence, Equation of tangent at point (1, 1) and with slope āˆ’1 is š‘¦āˆ’1=āˆ’1 (š‘„āˆ’1) š‘¦āˆ’1=āˆ’š‘„+1 š’š+š’™āˆ’šŸ = šŸŽ Also, Slope of Normal = (āˆ’1)/(š‘†š‘™š‘œš‘š‘’ š‘œš‘“ š‘”š‘Žš‘›š‘”š‘’š‘›š‘”) = (āˆ’1)/(āˆ’1) = 1 Thus, Equation of normal at point (1, 1) and with slope 1 is š‘¦ āˆ’ 1 = 1 (š‘„ āˆ’ 1) š‘¦ āˆ’ 1 = š‘„ āˆ’ 1 š‘¦ =š‘„ š’š āˆ’š’™=šŸŽ

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh is an IIT Kanpur graduate and has been teaching for 16+ years. At Teachoo, he breaks down Maths, Science and Computer Science into simple steps so students understand concepts deeply and score with confidence.

Many students prefer Teachoo Black for a smooth, ad-free learning experience.