Question 2 - Examples - Chapter 6 Class 12 Application of Derivatives

Last updated at December 16, 2024 by

Example 15 - Find point at which tangent to y = root 4x-3 - 1

Example 15 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 15 - Chapter 6 Class 12 Application of Derivatives - Part 3

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Question 2 Find the point at which the tangent to the curve š‘¦ = āˆš(4š‘„āˆ’3)āˆ’1 has its slope 2/3 . Given, Slope of the tangent to the curve is 2/3 We know that Slope of tangent = š’…š’š/š’…š’™ 2/3 = š‘‘š‘¦/š‘‘š‘„ š‘‘š‘¦/š‘‘š‘„ = 2/3 š‘‘(āˆš(4š‘„ āˆ’ 3) āˆ’ 1)/š‘‘š‘„ = 2/3 1/(2āˆš(4š‘„ āˆ’ 3)) Ɨ4āˆ’0 = 2/3 2/āˆš(4š‘„ āˆ’ 3) = 2/3 3 = āˆš(4š‘„āˆ’3) āˆš(šŸ’š’™āˆ’šŸ‘) = 3 Squaring both sides 4x āˆ’ 3 = 9 4x = 12 x = 3 Finding y for x = 3 š‘¦=āˆš(4š‘„āˆ’3) āˆ’ 1 =āˆš(12āˆ’3)āˆ’1 =āˆš9āˆ’1 =3āˆ’1 =šŸ Hence, the required point is (šŸ‘, šŸ)

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