Ex 6.3

Chapter 6 Class 12 Application of Derivatives
Serial order wise

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Ex 6.3, 27 (Method 1) The point on the curve π₯2=2π¦ which is nearest to the point (0, 5) is (A) (2 β2 ,4) (B) (2 β2,0) (C) (0, 0) (D) (2, 2) Let (β , π) be the point on the curve π₯2 = 2π¦ Where is nearest to the point (0, 5) Since (β, π) lie on the curve π₯2= 2π¦ β (β π) will satisfy the equation of curve π₯2=2π¦ β Putting π₯=β & y=π in equation β^2=2π We need to minimize the distance of a point (β ,π) from(0, 5) Let D be the distant between (β,π) & (0,5) D = β((0ββ)^2+(5βπ)^2 ) D = β(β^2+(5βπ^2 ) ) From (1) β^2=2π D = β(2π+(5βπ)^2 ) Diff w.r.t π ππ·/ππ=π(β(2π + (5 β π)^2 ))/ππ =1/(2β(2π + (5 β π)^2 )) Γπ(2π + (5 β π)^2 )/ππΎ =1/(2β(2π + (5 β π)^2 )) Γ [2+2(5βπ).π(5 β π)/ππΎ] Let D be the distant between (β,π) & (0,5) D = β((0ββ)^2+(5βπ)^2 ) D = β(β^2+(5βπ^2 ) ) From (1) β^2=2π D = β(2π+(5βπ)^2 ) We need to minimize D, but D has a square root Which will be difficult to differentiate Let Z = D2 Z = 2π+(5βπ)^2 Since D is positive, D is minimum if d2 is minimum So, we minimize Z = D2 Differentiating Z Z = 2π+(5βπ)^2 Diff w.r.t. k Zβ = π(2π + (5 β π)^2 )/ππ Zβ = 2 + 2 (5 β k) Γ (5 β k)β Zβ = 2 + 2 (5 β k) Γ (0 β 1) Zβ = 2 β 2 (5 β k) Zβ = 2 β 10 + 2k Zβ = β8 + 2k Putting Zβ = 0 β8 + 2k = 0 2k = 8 k = 8/2 = 4 Now, checking sign of (π^2 π)/(πβ^2 ) " " ππ/ππ=β8+2π Differentiating again w.r.t k (π^2 π)/(πβ^2 ) = 0+2 (π^2 π)/(πβ^2 ) = 2 β΄ (π^2 π)/(πβ^2 ) > 0 for k = 4 β΄ 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=β8 h=2β2 Hence, Required Point is (β,π)=(2β(2 ,) 4) Correct answer is A Ex 6.3, 27 (Method 2) The point on the curve π₯2= 2π¦ which is nearest to the point (0, 5) is (A) (2 β2,4) (B) (2 β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 (2β(2 ,) 4) Putting π₯=2β2 , & π¦=4 in π₯2=2π¦ β (2β2)^2=2(4) β 4 Γ 2 = 8 Which is true Thus, (2β2,4) lie on the curve Now, finding distance between (2β(2 ,) 4" " ) & (0 ,5) D = β((0β2β(2 ))^2+(5β4)^2 ) = β(8+1) = β9 = 3 Option 2 Point (2β(2 ,) 0) Putting π₯=2β2 & π¦=0 in π₯2=2π¦ (2β(2 ))^2=2(0) (4 Γ2)=0 8 = 0 Since 8 β 0 β (2β(2 ,) 0) is not the required point Option 3 Point (0, 0) Putting π₯=0 & π¦=0 in π₯2=2π¦ (0)^2=2(0) 0=0 β΄ (0 , 0) lie on the curve Now, Finding distance between (0, 0) πππ (0 , 5) D = β((0β0)^2+(5β0)^2 ) = β(0+5^2 ) = 5 Option 4 Point (2 ,2) Putting π₯=2 & π¦=2 in π₯2=2π¦ (2)^2=2(2) 4=4 β΄ (2, 2) lie on the curve Now, Finding distance between (2, 2) πππ (0 , 5) D = β((0β2)^2+(5β2)^2 ) = β((β2)^2+(3)^2 ) = β(4+9) = β13 Thus, Point (2β2,4) is on the curve π₯2=2π¦ & nearest to the point (0, 5) Hence correct answer is A