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


Transcript

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 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.