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Question 11 - Finding equation of tangent/normal when slope and curve are given - Chapter 6 Class 12 Application of Derivatives
Last updated at August 24, 2026 by Teachoo
Finding equation of tangent/normal when slope and curve are given
Chapter 6 Class 12 Application of Derivatives
Concept wise
- Finding rate of change
- To show increasing/decreasing in whole domain
- To show increasing/decreasing in intervals
- Find intervals of increasing/decreasing
- Finding slope of tangent/normal
- Finding point when tangent is parallel/ perpendicular
- Finding equation of tangent/normal when point and curve is given
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Finding equation of tangent/normal when slope and curve are given
- Finding approximate value of numbers
- Finding approximate value of function
- Finding approximate value- Statement questions
- Finding minimum and maximum values from graph
- Local maxima and minima
- Minima/ maxima (statement questions) - Number questions
- Minima/ maxima (statement questions) - Geometry questions
- Absolute minima/maxima
Transcript
Question 11 Find the equation of all lines having slope 2 which are tangents to the curve 𝑦=1/(𝑥 − 3) , 𝑥≠3.The Equation of Given Curve is : 𝑦=1/(𝑥 − 3) We know that Slope of tangent is 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥=𝑑(1/(𝑥 − 3))/𝑑𝑥 𝑑𝑦/𝑑𝑥=𝑑/𝑑𝑥 (𝑥−3)^(−1) 𝑑𝑦/𝑑𝑥=(−1) (𝑥−3)^(−1−1) . 𝑑(𝑥 − 3)/𝑑𝑥 𝑑𝑦/𝑑𝑥=−(𝑥−3)^(−2) 𝑑𝑦/𝑑𝑥=(−1)/(𝑥 − 3)^2 Given that slope = 2 Hence, 𝑑𝑦/𝑑𝑥 = 2 ∴ (−1)/(𝑥 − 3)^2 =2 −1=2(𝑥−3)^2 〖2(𝑥−3)〗^2=−1 (𝑥−3)^2=(−1)/( 2) We know that Square of any number is always positive So, (𝑥−3)^2>0 ∴ (𝑥−3)^2=(−1)/( 2) not possible Thus, No tangent to the Curve has Slope 2
