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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


Transcript

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.