Ex 6.5,27 - Chapter 6 Class 12 Application of Derivatives

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Ex 6.5,27 (Method 1) The point on the curve 𝑥2=2𝑦 which is nearest to the point (0, 5) is (A) (2 ﷮2﷯ ,4) (B) (2 ﷮2﷯,0) (C) (0, 0) (D) (2, 2) Let ℎ , 𝑘﷯ be the point on the curve 𝑥2 = 2𝑦 Where is nearest to the point 0 , 5﷯ Since ℎ , 𝑘﷯ lie on the curve 𝑥2 = 2𝑦 ⇒ ℎ , 𝑘﷯ will satisfy the equation of curve 𝑥2 = 2𝑦 ⇒ Putting 𝑥=ℎ & y=𝑘 in equation ℎ﷮2﷯=2𝑘 We need to minimize the distance of a point ℎ ,𝑘﷯from 0,5﷯ Let D be the distant between ℎ,𝑘﷯ & 0,5﷯ D = ﷮ 0−ℎ﷯﷮2﷯+ 5−𝑘﷯﷮2﷯﷯ D = ﷮ ℎ﷮2﷯+ 5− 𝑘﷮2﷯﷯﷯ D = ﷮2𝑘+ 5−𝑘﷯﷮2﷯﷯ Diff w.r.t 𝑘 𝑑𝐷﷮𝑑𝑘﷯= 𝑑 ﷮2𝑘 + 5 − 𝑘﷯﷮2﷯﷯﷯﷮𝑑𝑘﷯ = 1﷮2 ﷮2𝑘 + 5 − 𝑘﷯﷮2﷯﷯﷯ × 𝑑 2𝑘 + 5 − 𝑘﷯﷮2﷯﷯﷮𝑑𝐾﷯ = 1﷮2 ﷮2𝑘 + 5 − 𝑘﷯﷮2﷯﷯﷯ × 2+2 5−𝑘﷯. 𝑑 5 − 𝑘﷯﷮𝑑𝐾﷯﷯ = 1﷮2 ﷮2𝑘 + 5 − 𝑘﷯﷮2﷯﷯﷯ × 2+2 5−𝑘﷯ 0−1﷯﷯ = 1﷮2 ﷮2𝑘 + 5 − 𝑘﷯﷮2﷯﷯﷯ × 2−2 5−𝑘﷯﷯ = 2 − 2 5 − 𝑘﷯﷮2 ﷮2𝑘 + 5 − 𝑘﷯﷮2﷯﷯﷯ = 2 1 − 5 − 𝑘﷯﷯﷮2 ﷮2𝑘 + 5 − 𝑘﷯﷮2﷯﷯﷯ = 1 − 5 + 𝑘﷮ ﷮2𝑘 + 5 − 𝑘﷯﷮2﷯﷯﷯ = − 4 + 𝑘﷮ ﷮2𝑘 + 5 − 𝑘﷯﷮2﷯﷯﷯ Putting 𝑑𝐷﷮𝑑𝐾﷯=0 ⇒ − 4 + 𝑘﷮ ﷮2𝑘 + 5 − 𝑘﷯﷮2﷯﷯﷯=0 ⇒ – 4 + 𝑘=0 𝑘=4 Hence 𝑘=4 Thus, 𝑘 = 4 is point of minima D is minimum when 𝑘 = 4 Finding h From (1) h﷮2﷯=2𝑘 h﷮2﷯=2 4﷯ h﷮2﷯=8 ℎ= ﷮8﷯ ℎ=2 ﷮2﷯ Hence, Required Point is ℎ,𝑘﷯= 2 ﷮2 ,﷯4﷯ Correct answer is A Ex 6.5,27 (Method 2) The point on the curve 𝑥2 = 2𝑦 which is nearest to the point (0, 5) is (A) (2 ﷮2﷯,4) (B) (2 ﷮2﷯,0) (C) (0, 0) (D) (2, 2) Since points given lie on the curve, it will statisfy equation of curve Option 1 Point is 2 ﷮2 ,﷯4﷯ Putting 𝑥=2 ﷮2﷯ , & 𝑦=4 in 𝑥2=2𝑦 ⇒ 2 ﷮2﷯﷯﷮2﷯=2 4﷯ ⇒ 4 × 2 = 8 Which is true Thus, 2 ﷮2﷯,4﷯ lie on the curve Now, finding distance between 2 ﷮2 ,﷯4 ﷯ & 0 ,5﷯ D = ﷮ 0−2 ﷮2 ﷯﷯﷮2﷯+ 5−4﷯﷮2﷯﷯ = ﷮8+1﷯ = ﷮9﷯ = 3 Option 2 Point 2 ﷮2 ,﷯0﷯ Putting 𝑥=2 ﷮2﷯ & 𝑦=0 in 𝑥2=2𝑦 2 ﷮2 ﷯﷯﷮2﷯=2 0﷯ 4 ×2﷯=0 8 = 0 Since 8 ≠0 ⇒ 2 ﷮2 ,﷯0﷯ is not the required point Option 3 Point 0 ,0﷯ Putting 𝑥=0 & 𝑦=0 in 𝑥2=2𝑦 0﷯﷮2﷯=2 0﷯ 0=0 ∴ 0 , 0﷯ lie on the curve Now, Finding distance between 0 , 0﷯ 𝑎𝑛𝑑 0 , 5﷯ D = ﷮ 0−0﷯﷮2﷯+ 5−0﷯﷮2﷯﷯ = ﷮0+ 5﷮2﷯﷯ = 5 Option 4 Point 2 ,2﷯ Putting 𝑥=2 & 𝑦=2 in 𝑥2=2𝑦 2﷯﷮2﷯=2 2﷯ 4=4 ∴ 2, 2﷯ lie on the curve Now, Finding distance between 2, 2﷯ 𝑎𝑛𝑑 0 , 5﷯ D = ﷮ 0−2﷯﷮2﷯+ 5−2﷯﷮2﷯﷯ = ﷮ −2﷯﷮2﷯+ 3﷯﷮2﷯﷯ = ﷮4+9﷯ = ﷮13﷯ Thus, Point 2 ﷮2 ,﷯4﷯ is on the curve 𝑥2=2𝑦 & nearest to the point 0 , 5﷯ Hence correct answer is A