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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Ex 6.3, 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 ) ) From (1) ℎ^2=2𝑘 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] Let D be the distant between (ℎ,𝑘) & (0,5) D = √((0−ℎ)^2+(5−𝑘)^2 ) D = √(ℎ^2+(5−𝑘^2 ) ) From (1) ℎ^2=2𝑘 D = √(2𝑘+(5−𝑘)^2 ) We need to minimize D, but D has a square root Which will be difficult to differentiate Let Z = D2 Z = 2𝑘+(5−𝑘)^2 Since D is positive, D is minimum if d2 is minimum So, we minimize Z = D2 Differentiating Z Z = 2𝑘+(5−𝑘)^2 Diff w.r.t. k Z’ = 𝑑(2𝑘 + (5 − 𝑘)^2 )/𝑑𝑘 Z’ = 2 + 2 (5 − k) × (5 − k)’ Z’ = 2 + 2 (5 − k) × (0 − 1) Z’ = 2 − 2 (5 − k) Z’ = 2 − 10 + 2k Z’ = −8 + 2k Putting Z’ = 0 −8 + 2k = 0 2k = 8 k = 8/2 = 4 Now, checking sign of (𝑑^2 𝑍)/(𝑑ℎ^2 ) " " 𝑑𝑍/𝑑𝑘=−8+2𝑘 Differentiating again w.r.t k (𝑑^2 𝑍)/(𝑑ℎ^2 ) = 0+2 (𝑑^2 𝑍)/(𝑑ℎ^2 ) = 2 ∴ (𝑑^2 𝑍)/(𝑑ℎ^2 ) > 0 for k = 4 ∴ Z is minimum when k = 4 Thus, D is Minimum at 𝑘= 4 Finding h From (1) h^2=2𝑘 h^2=2(4) h=√8 h=2√2 Hence, Required Point is (ℎ,𝑘)=(2√(2 ,) 4) Correct answer is A Ex 6.3, 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 satisfy 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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.