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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Example 23 Find the shortest distance of the point (0, c) from the parabola 𝑦=𝑥2, where 0 ≤ c ≤ 5. Let (ℎ ,𝑘) be any point on parabola 𝑦=𝑥2 Let D be required Distance between (ℎ , 𝑘) & (0 , 𝑐) D = √((0−ℎ)^2+(𝑐 −𝑘)^2 ) D = √((−ℎ)^2+(𝑐 −𝑘)^2 ) D = √(𝒉^𝟐+(𝒄 −𝒌)^𝟐 ) Also, Since point (ℎ , 𝑘) is on the parabola 𝑦=𝑥2 (𝒉 , 𝒌) will satisfy the equation of parabola Putting 𝑥=ℎ , 𝑦=𝑘 in equation 𝒌=𝒉^𝟐 Putting value of 𝑘=ℎ^2 D = √(ℎ^2+(𝑐 −𝑘)^2 ) D = √(𝒌+(𝒄−𝒌)^𝟐 ) We need to minimize D, but D has a square root Which will be difficult to differentiate Let Z = D2 Z = 𝑘+(𝑐−𝑘)^2 Since D is positive, D is minimum if D2 is minimum So, we minimize Z = D2 Differentiating Z Z =𝑘+(𝑐−𝑘)^2 Differentiating w.r.t. k Z’ = 𝑑(𝑘 + (𝑐 − 𝑘)^2 )/𝑑𝑘 Z’ = 1 + 2 (c − k) × (c − k)’ Z’ = 1 + 2 (c − k) × (0 − 1) Z’ = 1 − 2 (c − k) Z’ = 1 − 2c − 2k Putting Z’ = 0 1 − 2c − 2k = 0 2k = 2c − 1 k = (𝟐𝒄 − 𝟏)/𝟐 Now, checking sign of 𝒁^′′ " " 𝑑𝑍/𝑑𝑘=4𝑘−2𝑐 Differentiating again w.r.t k (𝑑^2 𝑍)/(𝑑ℎ^2 ) = 4 −0 (𝒅^𝟐 𝒁)/(𝒅𝒉^𝟐 ) = 𝟒 Since 𝐙^′′ > 0 for k = (2𝑐 − 1)/2 ∴ Z is minimum when k = (2𝑐 − 1)/2 Thus, D is Minimum at 𝒌=(𝟐𝒄 − 𝟏)/𝟐 Finding Minimum value of D D = √(𝑘+(𝑐−𝑘)^2 ) Putting 𝑘=(2𝑐 − 1)/2 D = √(((2𝑐 − 1)/2)+(𝑐−((2𝑐 − 1)/2))^2 ) D = √(((2𝑐 − 1)/2)+((2𝑐 − 2𝑐 − 1)/2)^2 ) D = √(((2𝑐 − 1)/2)+((−1)/2)^2 ) D = √(((2𝑐 − 1)/2)+1/4) D = √(𝑐−1/2+1/4) D = √(𝑐−1/4) D = √(4𝑐 − 1)/2 Hence, shortest distance is √(𝟒𝒄 − 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.