# Ex 6.3, 27 (MCQ) - Chapter 6 Class 12 Application of Derivatives

Last updated at April 16, 2024 by Teachoo

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Ex 6.3, 27 (MCQ) You are here

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Ex 6.3,29 (MCQ)

Last updated at April 16, 2024 by Teachoo

Ex 6.3, 27 (Method 1) The point on the curve š„2=2š¦ which is nearest to the point (0, 5) is (A) (2 ā2 ,4) (B) (2 ā2,0) (C) (0, 0) (D) (2, 2) Let (ā , š) be the point on the curve š„2 = 2š¦ Where is nearest to the point (0, 5) Since (ā, š) lie on the curve š„2= 2š¦ ā (ā š) will satisfy the equation of curve š„2=2š¦ ā Putting š„=ā & y=š in equation ā^2=2š We need to minimize the distance of a point (ā ,š) from(0, 5) Let D be the distant between (ā,š) & (0,5) D = ā((0āā)^2+(5āš)^2 ) D = ā(ā^2+(5āš^2 ) ) From (1) ā^2=2š D = ā(2š+(5āš)^2 ) Diff w.r.t š šš·/šš=š(ā(2š + (5 ā š)^2 ))/šš =1/(2ā(2š + (5 ā š)^2 )) Ćš(2š + (5 ā š)^2 )/šš¾ =1/(2ā(2š + (5 ā š)^2 )) Ć [2+2(5āš).š(5 ā š)/šš¾] Let D be the distant between (ā,š) & (0,5) D = ā((0āā)^2+(5āš)^2 ) D = ā(ā^2+(5āš^2 ) ) From (1) ā^2=2š D = ā(2š+(5āš)^2 ) We need to minimize D, but D has a square root Which will be difficult to differentiate Let Z = D2 Z = 2š+(5āš)^2 Since D is positive, D is minimum if d2 is minimum So, we minimize Z = D2 Differentiating Z Z = 2š+(5āš)^2 Diff w.r.t. k Zā = š(2š + (5 ā š)^2 )/šš Zā = 2 + 2 (5 ā k) Ć (5 ā k)ā Zā = 2 + 2 (5 ā k) Ć (0 ā 1) Zā = 2 ā 2 (5 ā k) Zā = 2 ā 10 + 2k Zā = ā8 + 2k Putting Zā = 0 ā8 + 2k = 0 2k = 8 k = 8/2 = 4 Now, checking sign of (š^2 š)/(šā^2 ) " " šš/šš=ā8+2š Differentiating again w.r.t k (š^2 š)/(šā^2 ) = 0+2 (š^2 š)/(šā^2 ) = 2 ā“ (š^2 š)/(šā^2 ) > 0 for k = 4 ā“ Z is minimum when k = 4 Thus, D is Minimum at š= 4 Finding h From (1) h^2=2š h^2=2(4) h=ā8 h=2ā2 Hence, Required Point is (ā,š)=(2ā(2 ,) 4) Correct answer is A Ex 6.3, 27 (Method 2) The point on the curve š„2= 2š¦ which is nearest to the point (0, 5) is (A) (2 ā2,4) (B) (2 ā2,0) (C) (0, 0) (D) (2, 2)Since points given lie on the curve, it will satisfy equation of curve Option 1 Point is (2ā(2 ,) 4) Putting š„=2ā2 , & š¦=4 in š„2=2š¦ ā (2ā2)^2=2(4) ā 4 Ć 2 = 8 Which is true Thus, (2ā2,4) lie on the curve Now, finding distance between (2ā(2 ,) 4" " ) & (0 ,5) D = ā((0ā2ā(2 ))^2+(5ā4)^2 ) = ā(8+1) = ā9 = 3 Option 2 Point (2ā(2 ,) 0) Putting š„=2ā2 & š¦=0 in š„2=2š¦ (2ā(2 ))^2=2(0) (4 Ć2)=0 8 = 0 Since 8 ā 0 ā (2ā(2 ,) 0) is not the required point Option 3 Point (0, 0) Putting š„=0 & š¦=0 in š„2=2š¦ (0)^2=2(0) 0=0 ā“ (0 , 0) lie on the curve Now, Finding distance between (0, 0) ššš (0 , 5) D = ā((0ā0)^2+(5ā0)^2 ) = ā(0+5^2 ) = 5 Option 4 Point (2 ,2) Putting š„=2 & š¦=2 in š„2=2š¦ (2)^2=2(2) 4=4 ā“ (2, 2) lie on the curve Now, Finding distance between (2, 2) ššš (0 , 5) D = ā((0ā2)^2+(5ā2)^2 ) = ā((ā2)^2+(3)^2 ) = ā(4+9) = ā13 Thus, Point (2ā2,4) is on the curve š„2=2š¦ & nearest to the point (0, 5) Hence correct answer is A