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Question 27 (MCQ) - Finding slope of tangent/normal - Chapter 6 Class 12 Application of Derivatives
Last updated at August 13, 2026 by Teachoo
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
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Finding slope of tangent/normal
- Finding point when tangent is parallel/ perpendicular
- Finding equation of tangent/normal when point and curve is given
- 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
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Transcript
Question 27 The line π¦=π₯+1 is a tangent to the curve π¦2=4π₯ at the point (A) (1, 2) (B) (2, 1) (C) (1, β 2) (D) (β 1, 2)Given Curve is π¦^2=4π₯ Differentiating w.r.t. π₯ π(π¦^2 )/ππ₯=π(4π₯)/ππ₯ π(π¦^2 )/ππ¦ Γ ππ¦/ππ₯=4 2π¦ Γ ππ¦/ππ₯=4 ππ¦/ππ₯=4/2π¦ ππ¦/ππ₯=2/π¦ Given line is π¦=π₯+1 The Above line is of the form π¦=ππ₯+π when m is slope of line Slope of line π¦=π₯+1 is 1 Now Slope of tangent = Slope of line ππ¦/ππ₯=1 2/π¦=1 2=π¦ π¦=2 Finding x when π¦=2 π¦^2=4π₯ (2)^2=4π₯ 4=4π₯ 4/4=π₯ π₯=1 Hence the Required point is (x, y) = (1 , 2) Correct Answer is (A)
