# Ex 6.5,27

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 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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Ex 6.5,27 You are here

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Chapter 6 Class 12 Application of Derivatives

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.