Check sibling questions

Example 35 - Find shortest distance of (0, c) from parabola

Example 35 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 35 - Chapter 6 Class 12 Application of Derivatives - Part 3
Example 35 - Chapter 6 Class 12 Application of Derivatives - Part 4
Example 35 - Chapter 6 Class 12 Application of Derivatives - Part 5
Example 35 - Chapter 6 Class 12 Application of Derivatives - Part 6

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Example 35 Find the shortest distance of the point (0, c) from the parabola 𝑦=π‘₯2, where 0 ≀ c ≀ 5. Let (β„Ž ,π‘˜) be any point on parabola 𝑦=π‘₯2 Let D be required Distance between (β„Ž , π‘˜) & (0 , 𝑐) D = √((0βˆ’β„Ž)^2+(𝑐 βˆ’π‘˜)^2 ) D = √((βˆ’β„Ž)^2+(𝑐 βˆ’π‘˜)^2 ) D = √(𝒉^𝟐+(𝒄 βˆ’π’Œ)^𝟐 ) Distance between two (π‘₯1,𝑦1) & (π‘₯2 , 𝑦2) point is 𝑑= √((π‘₯2βˆ’π‘¦1)^2+(π‘₯2 βˆ’π‘¦1)^2 ) Also, Since point (β„Ž , π‘˜) is on the parabola 𝑦=π‘₯2 (𝒉 , π’Œ) will satisfy the equation of parabola Putting π‘₯=β„Ž , 𝑦=π‘˜ in equation π’Œ=𝒉^𝟐 Putting value of π‘˜=β„Ž^2 D = √(β„Ž^2+(𝑐 βˆ’π‘˜)^2 ) D = √(π’Œ+(π’„βˆ’π’Œ)^𝟐 ) We need to minimize D, but D has a square root Which will be difficult to differentiate Let Z = D2 Z = π‘˜+(π‘βˆ’π‘˜)^2 Since D is positive, D is minimum if D2 is minimum So, we minimize Z = D2 Differentiating Z Z =π‘˜+(π‘βˆ’π‘˜)^2 Differentiating w.r.t. k Z’ = 𝑑(π‘˜ + (𝑐 βˆ’ π‘˜)^2 )/π‘‘π‘˜ Z’ = 1 + 2 (c βˆ’ k) Γ— (c βˆ’ k)’ Z’ = 1 + 2 (c βˆ’ k) Γ— (0 βˆ’ 1) Z’ = 1 βˆ’ 2 (c βˆ’ k) Z’ = 1 βˆ’ 2c βˆ’ 2k Putting Z’ = 0 1 βˆ’ 2c βˆ’ 2k = 0 2k = 2c βˆ’ 1 k = (πŸπ’„ βˆ’ 𝟏)/𝟐 Now, checking sign of 𝒁^β€²β€² " " 𝑑𝑍/π‘‘π‘˜=4π‘˜βˆ’2𝑐 Differentiating again w.r.t k (𝑑^2 𝑍)/(π‘‘β„Ž^2 ) = 4 βˆ’0 (𝒅^𝟐 𝒁)/(𝒅𝒉^𝟐 ) = πŸ’ Since 𝐙^β€²β€² > 0 for k = (2𝑐 βˆ’ 1)/2 ∴ Z is minimum when k = (2𝑐 βˆ’ 1)/2 Thus, D is Minimum at π’Œ=(πŸπ’„ βˆ’ 𝟏)/𝟐 Finding Minimum value of D D = √(π‘˜+(π‘βˆ’π‘˜)^2 ) Putting π‘˜=(2𝑐 βˆ’ 1)/2 D = √(((2𝑐 βˆ’ 1)/2)+(π‘βˆ’((2𝑐 βˆ’ 1)/2))^2 ) D = √(((2𝑐 βˆ’ 1)/2)+((2𝑐 βˆ’ 2𝑐 βˆ’ 1)/2)^2 ) D = √(((2𝑐 βˆ’ 1)/2)+((βˆ’1)/2)^2 ) D = √(((2𝑐 βˆ’ 1)/2)+1/4) D = √(π‘βˆ’1/2+1/4) D = √(π‘βˆ’1/4) D = √(4𝑐 βˆ’ 1)/2 Hence, shortest distance is √(πŸ’π’„ βˆ’ 𝟏)/𝟐

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Davneet Singh

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.