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

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

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 ℎ2=2𝑘 We need to minimize the distance of a point ℎ ,𝑘from 0,5 Let D be the distant between ℎ,𝑘 & 0,5 D = 0−ℎ2+ 5−𝑘2 D = ℎ2+ 5− 𝑘2 D = 2𝑘+ 5−𝑘2 Diff w.r.t 𝑘 𝑑𝐷𝑑𝑘= 𝑑 2𝑘 + 5 − 𝑘2𝑑𝑘 = 12 2𝑘 + 5 − 𝑘2 × 𝑑 2𝑘 + 5 − 𝑘2𝑑𝐾 = 12 2𝑘 + 5 − 𝑘2 × 2+2 5−𝑘. 𝑑 5 − 𝑘𝑑𝐾 = 12 2𝑘 + 5 − 𝑘2 × 2+2 5−𝑘 0−1 = 12 2𝑘 + 5 − 𝑘2 × 2−2 5−𝑘 = 2 − 2 5 − 𝑘2 2𝑘 + 5 − 𝑘2 = 2 1 − 5 − 𝑘2 2𝑘 + 5 − 𝑘2 = 1 − 5 + 𝑘 2𝑘 + 5 − 𝑘2 = − 4 + 𝑘 2𝑘 + 5 − 𝑘2 Putting 𝑑𝐷𝑑𝐾=0 ⇒ − 4 + 𝑘 2𝑘 + 5 − 𝑘2=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) h2=2𝑘 h2=2 4 h2=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 22=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+ 5−42 = 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 02=2 0 0=0 ∴ 0 , 0 lie on the curve Now, Finding distance between 0 , 0 𝑎𝑛𝑑 0 , 5 D = 0−02+ 5−02 = 0+ 52 = 5 Option 4 Point 2 ,2 Putting 𝑥=2 & 𝑦=2 in 𝑥2=2𝑦 22=2 2 4=4 ∴ 2, 2 lie on the curve Now, Finding distance between 2, 2 𝑎𝑛𝑑 0 , 5 D = 0−22+ 5−22 = −22+ 32 = 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

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Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.