Misc 21 - Line y = mx + 1 is tangent to y2 = 4x if value of m is - Finding slope of tangent/normal

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  1. Chapter 6 Class 12 Application of Derivatives
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Misc 21 The line 𝑦=𝑚𝑥+1 is a tangent to the curve ﷐𝑦﷮2﷯=4𝑥 if the value of 𝑚 is (A) 1 (B) 2 (C) 3 (D) ﷐1﷮2﷯ Let ﷐ℎ , 𝑘﷯ be the point at which tangent is to be taken & Given Equation of tangent 𝑦=𝑚𝑥+1 & Curve is ﷐𝑦﷮2﷯=4𝑥 We know that Slope of tangent to the Curve is ﷐𝑑𝑦﷮𝑑𝑥﷯ ﷐𝑦﷮2﷯=4𝑥 Differentiating w.r.t. 𝑥 ﷐𝑑﷐﷐𝑦﷮2﷯﷯﷮𝑑𝑥﷯=﷐𝑑﷐4𝑥﷯﷮𝑑𝑥﷯ ﷐𝑑﷐﷐𝑦﷮2﷯﷯﷮𝑑𝑥﷯ × ﷐𝑑𝑦﷮𝑑𝑦﷯=4 ﷐𝑑﷐﷐𝑦﷮2﷯﷯﷮𝑑𝑦﷯ × ﷐𝑑𝑦﷮𝑑𝑥﷯=4 2𝑦 ×﷐𝑑𝑦﷮𝑑𝑥﷯=4 ﷐𝑑𝑦﷮𝑑𝑥﷯=﷐2﷮𝑦﷯ Since tangent to be taken from ﷐ℎ , 𝑘﷯ So, Slope of tangent at ﷐ℎ , 𝑘﷯ is ﷐﷐﷐𝑑𝑦﷮𝑑𝑥﷯﷯﷮﷐ℎ , 𝑘﷯﷯=﷐2﷮𝑘﷯ But Given Equation of tangent 𝑦=𝑚𝑥+1 Hence Slope of tangent = m ⇒ 𝑚 =﷐2﷮𝑘﷯ Now, Point ﷐ℎ , 𝑘﷯ lie on the Curve ﷐𝑦﷮2﷯=4𝑥 ⇒﷐ℎ , 𝑘﷯ will Satisfies the Equation of Curve Putting 𝑥=ℎ , 𝑦=𝑘 in equation ⇒ ﷐𝑘﷮2﷯=4ℎ Also, Point ﷐ℎ , 𝑘﷯ lie on the tangent ⇒ ﷐ℎ , 𝑘﷯ will satisfies the Equation of tangent 𝑦=𝑚𝑥+1 Putting 𝑥=ℎ , 𝑦=𝑘 in equation 𝑘=𝑚ℎ+1 From (1) Putting Value of 𝑚=﷐2﷮𝑘﷯ 𝑘=﷐2﷮𝑘﷯ ×ℎ+1 𝑘=﷐2ℎ + 𝑘﷮𝑘﷯ ﷐𝑘﷮2﷯=2ℎ+𝑘 ﷐𝑘﷮2﷯−𝑘=2ℎ Now Our Equation are ﷐𝑘﷮2﷯=4ℎ …(2) ﷐𝑘﷮2﷯−𝑘=2ℎ …(3) Solving (2) ﷐𝑘﷮2﷯=4ℎ ﷐𝑘﷮2﷯=2﷐2ℎ﷯ Putting 2ℎ=﷐𝑘﷮2﷯−𝑘 from (3) ﷐𝑘﷮2﷯=2﷐﷐𝑘﷮2﷯−𝑘﷯ ﷐𝑘﷮2﷯=2﷐𝑘﷮2﷯−2𝑘 0=2﷐𝑘﷮2﷯−2𝑘−﷐𝑘﷮2﷯ 2﷐𝑘﷮2﷯−﷐𝑘﷮2﷯−2𝑘=0 ﷐𝑘﷮2﷯−2𝑘=0 𝑘﷐𝑘−2﷯=0 Hence 𝑘=0 , 2 Since 𝑚=𕔴 not in the Option So, m = 1 Hence Correct Answer is (A)

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