# Ex 6.5, 27 (MCQ) - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at Aug. 9, 2021 by Teachoo

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Ex 6.5, 27 (MCQ) You are here

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Ex 6.5,29 (MCQ)

Last updated at Aug. 9, 2021 by Teachoo

Ex 6.5, 27 (Method 1) The point on the curve π₯2=2π¦ which is nearest to the point (0, 5) is (A) (2 β2 ,4) (B) (2 β2,0) (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((0ββ)^2+(5βπ)^2 ) D = β(β^2+(5βπ^2 ) ) From (1) β^2=2π D = β(2π+(5βπ)^2 ) Diff w.r.t π ππ·/ππ=π(β(2π + (5 β π)^2 ))/ππ =1/(2β(2π + (5 β π)^2 )) Γπ(2π + (5 β π)^2 )/ππΎ =1/(2β(2π + (5 β π)^2 )) Γ [2+2(5βπ).π(5 β π)/ππΎ] Let D be the distant between (β,π) & (0,5) D = β((0ββ)^2+(5βπ)^2 ) D = β(β^2+(5βπ^2 ) ) From (1) β^2=2π D = β(2π+(5βπ)^2 ) We need to minimize D, but D has a square root Which will be difficult to differentiate Let Z = D2 Z = 2π+(5βπ)^2 Since D is positive, D is minimum if d2 is minimum So, we minimize Z = D2 Differentiating Z Z = 2π+(5βπ)^2 Diff w.r.t. k Zβ = π(2π + (5 β π)^2 )/ππ 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 = β8 + 2k Putting Zβ = 0 β8 + 2k = 0 2k = 8 k = 8/2 = 4 Now, checking sign of (π^2 π)/(πβ^2 ) " " ππ/ππ=β8+2π Differentiating again w.r.t k (π^2 π)/(πβ^2 ) = 0+2 (π^2 π)/(πβ^2 ) = 2 β΄ (π^2 π)/(πβ^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=β8 h=2β2 Hence, Required Point is (β,π)=(2β(2 ,) 4) Correct answer is A Ex 6.5, 27 (Method 2) The point on the curve π₯2= 2π¦ which is nearest to the point (0, 5) is (A) (2 β2,4) (B) (2 β2,0) (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₯=2β2 , & π¦=4 in π₯2=2π¦ β (2β2)^2=2(4) β 4 Γ 2 = 8 Which is true Thus, (2β2,4) lie on the curve Now, finding distance between (2β(2 ,) 4" " ) & (0 ,5) D = β((0β2β(2 ))^2+(5β4)^2 ) = β(8+1) = β9 = 3 Option 2 Point (2β(2 ,) 0) Putting π₯=2β2 & π¦=0 in π₯2=2π¦ (2β(2 ))^2=2(0) (4 Γ2)=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) πππ (0 , 5) D = β((0β0)^2+(5β0)^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) πππ (0 , 5) D = β((0β2)^2+(5β2)^2 ) = β((β2)^2+(3)^2 ) = β(4+9) = β13 Thus, Point (2β2,4) is on the curve π₯2=2π¦ & nearest to the point (0, 5) Hence correct answer is A