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Ex 6.5, 27 - The point on curve x^2 = 2y which is nearest to (0, 5) is

Ex 6.5, 27 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.5, 27 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.5, 27 - Chapter 6 Class 12 Application of Derivatives - Part 4 Ex 6.5, 27 - Chapter 6 Class 12 Application of Derivatives - Part 5 Ex 6.5, 27 - Chapter 6 Class 12 Application of Derivatives - Part 6 Ex 6.5, 27 - Chapter 6 Class 12 Application of Derivatives - Part 7 Ex 6.5, 27 - Chapter 6 Class 12 Application of Derivatives - Part 8 Ex 6.5, 27 - Chapter 6 Class 12 Application of Derivatives - Part 9 Ex 6.5, 27 - Chapter 6 Class 12 Application of Derivatives - Part 10 Ex 6.5, 27 - Chapter 6 Class 12 Application of Derivatives - Part 11

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Ex 6.5, 27 (Method 1) The point on the curve π‘₯2=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 (β„Ž , π‘˜) be the point on the curve π‘₯2 = 2𝑦 Where is nearest to the point (0, 5) Since (β„Ž, π‘˜) lie on the curve π‘₯2= 2𝑦 β‡’ (β„Ž π‘˜) will satisfy the equation of curve π‘₯2=2𝑦 β‡’ Putting π‘₯=β„Ž & y=π‘˜ in equation β„Ž^2=2π‘˜ We need to minimize the distance of a point (β„Ž ,π‘˜) from(0, 5) Let D be the distant between (β„Ž,π‘˜) & (0,5) D = √((0βˆ’β„Ž)^2+(5βˆ’π‘˜)^2 ) D = √(β„Ž^2+(5βˆ’π‘˜^2 ) ) From (1) β„Ž^2=2π‘˜ D = √(2π‘˜+(5βˆ’π‘˜)^2 ) Diff w.r.t π‘˜ 𝑑𝐷/π‘‘π‘˜=𝑑(√(2π‘˜ + (5 βˆ’ π‘˜)^2 ))/π‘‘π‘˜ =1/(2√(2π‘˜ + (5 βˆ’ π‘˜)^2 )) ×𝑑(2π‘˜ + (5 βˆ’ π‘˜)^2 )/𝑑𝐾 =1/(2√(2π‘˜ + (5 βˆ’ π‘˜)^2 )) Γ— [2+2(5βˆ’π‘˜).𝑑(5 βˆ’ π‘˜)/𝑑𝐾] Let D be the distant between (β„Ž,π‘˜) & (0,5) D = √((0βˆ’β„Ž)^2+(5βˆ’π‘˜)^2 ) D = √(β„Ž^2+(5βˆ’π‘˜^2 ) ) From (1) β„Ž^2=2π‘˜ D = √(2π‘˜+(5βˆ’π‘˜)^2 ) We need to minimize D, but D has a square root Which will be difficult to differentiate Let Z = D2 Z = 2π‘˜+(5βˆ’π‘˜)^2 Since D is positive, D is minimum if d2 is minimum So, we minimize Z = D2 Differentiating Z Z = 2π‘˜+(5βˆ’π‘˜)^2 Diff w.r.t. k Z’ = 𝑑(2π‘˜ + (5 βˆ’ π‘˜)^2 )/π‘‘π‘˜ Z’ = 2 + 2 (5 βˆ’ k) Γ— (5 βˆ’ k)’ Z’ = 2 + 2 (5 βˆ’ k) Γ— (0 βˆ’ 1) Z’ = 2 βˆ’ 2 (5 βˆ’ k) Z’ = 2 βˆ’ 10 + 2k Z’ = βˆ’8 + 2k Putting Z’ = 0 βˆ’8 + 2k = 0 2k = 8 k = 8/2 = 4 Now, checking sign of (𝑑^2 𝑍)/(π‘‘β„Ž^2 ) " " 𝑑𝑍/π‘‘π‘˜=βˆ’8+2π‘˜ Differentiating again w.r.t k (𝑑^2 𝑍)/(π‘‘β„Ž^2 ) = 0+2 (𝑑^2 𝑍)/(π‘‘β„Ž^2 ) = 2 ∴ (𝑑^2 𝑍)/(π‘‘β„Ž^2 ) > 0 for k = 4 ∴ Z is minimum when k = 4 Thus, D is Minimum at π‘˜= 4 Finding h From (1) h^2=2π‘˜ h^2=2(4) h=√8 h=2√2 Hence, Required Point is (β„Ž,π‘˜)=(2√(2 ,) 4) Correct answer is A Ex 6.5, 27 (Method 2) The point on the curve π‘₯2= 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 π‘₯=2√2 , & 𝑦=4 in π‘₯2=2𝑦 β‡’ (2√2)^2=2(4) β‡’ 4 Γ— 2 = 8 Which is true Thus, (2√2,4) lie on the curve Now, finding distance between (2√(2 ,) 4" " ) & (0 ,5) D = √((0βˆ’2√(2 ))^2+(5βˆ’4)^2 ) = √(8+1) = √9 = 3 Option 2 Point (2√(2 ,) 0) Putting π‘₯=2√2 & 𝑦=0 in π‘₯2=2𝑦 (2√(2 ))^2=2(0) (4 Γ—2)=0 8 = 0 Since 8 β‰ 0 β‡’ (2√(2 ,) 0) is not the required point Option 3 Point (0, 0) Putting π‘₯=0 & 𝑦=0 in π‘₯2=2𝑦 (0)^2=2(0) 0=0 ∴ (0 , 0) lie on the curve Now, Finding distance between (0, 0) π‘Žπ‘›π‘‘ (0 , 5) D = √((0βˆ’0)^2+(5βˆ’0)^2 ) = √(0+5^2 ) = 5 Option 4 Point (2 ,2) Putting π‘₯=2 & 𝑦=2 in π‘₯2=2𝑦 (2)^2=2(2) 4=4 ∴ (2, 2) lie on the curve Now, Finding distance between (2, 2) π‘Žπ‘›π‘‘ (0 , 5) D = √((0βˆ’2)^2+(5βˆ’2)^2 ) = √((βˆ’2)^2+(3)^2 ) = √(4+9) = √13 Thus, Point (2√2,4) is on the curve π‘₯2=2𝑦 & nearest to the point (0, 5) Hence correct answer is A

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.