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

Slide16.JPG
Slide17.JPG Slide18.JPG Slide19.JPG Slide20.JPG Slide21.JPG

  1. Class 12
  2. Important Question for exams Class 12
Ask Download

Transcript

Example 35 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 the parabola 𝑦=𝑥2 Let D be the Required Distance between ﷐ℎ , 𝑘﷯ & ﷐𝑜 , 𝑐﷯ D = ﷐﷮﷐﷐0−ℎ﷯﷮2﷯+﷐﷐𝑐 −𝑘﷯﷮2﷯﷯ D = ﷐﷮﷐﷐−ℎ﷯﷮2﷯+﷐﷐𝑐 −𝑘﷯﷮2﷯﷯ D = ﷐﷮﷐ℎ﷮2﷯+﷐﷐𝑐 −𝑘﷯﷮2﷯﷯ 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﷯ in (1) D = ﷐﷮𝑘+﷐﷐𝑐−𝑘﷯﷮2﷯﷯ We need to minimize distance D & find point ﷐ℎ , 𝑘﷯ Finding D’﷐𝑘﷯ D = ﷐﷮𝑘+﷐﷐𝑐−𝑘﷯﷮2﷯﷯ Diff w.r.t k D’ ﷐𝑘﷯=﷐1﷮2﷐﷮𝑘 + ﷐﷐𝑐 − 𝑘﷯﷮2﷯﷯﷯ × ﷐𝑑﷐𝑘 + ﷐﷐𝑐 − 𝑘﷯﷮2﷯﷯﷮𝑑𝑘﷯ = ﷐1﷮2﷐﷮𝑘 + ﷐﷐𝑐 − 𝑘﷯﷮2﷯﷯﷯ × ﷐1+2﷐𝑐−𝑘﷯.﷐𝑑﷐𝑐 − 𝑘﷯﷮𝑑𝑘﷯﷯ = ﷐1﷮2﷐﷮𝑘 + ﷐﷐𝑐 − 𝑘﷯﷮2﷯﷯﷯ × ﷐1+2﷐𝑐−𝑘﷯.﷐0−1﷯﷯ = ﷐1﷮2﷐﷮𝑘 + ﷐﷐𝑐 − 𝑘﷯﷮2﷯﷯﷯ × ﷐1+2﷐𝑐−𝑘﷯.﷐−1﷯﷯ = ﷐1﷮2﷐﷮𝑘 + ﷐﷐𝑐 − 𝑘﷯﷮2﷯﷯﷯ × ﷐1−2﷐𝑐−𝑘﷯﷯ = ﷐1 − 2﷐𝑐 − 𝑘﷯﷮2﷐﷮𝑘 + ﷐﷐𝑐 − 𝑘﷯﷮2﷯﷯﷯ Putting D’ ﷐𝑘﷯=0 ﷐1 − 2﷐𝑐 − 𝑘﷯﷮2﷐﷮𝑘 + ﷐﷐𝑐 − 𝑘﷯﷮2﷯﷯﷯=0 1 – 2﷐𝑐−𝑘﷯=0 2﷐𝑐−𝑘﷯=1 c – k = ﷐1﷮2﷯ c – ﷐1﷮2﷯ = k 𝑘=c−﷐1﷮2﷯ Now, finding sign of D’(k) for values less than & greater than 𝑘=c−﷐1﷮2﷯ Thus, D is Minimum at 𝑘=c −﷐1 ﷮2﷯ Now, From (2) 𝑘=﷐ℎ﷮2﷯ ﷐ℎ﷮2﷯ = k ﷐ℎ﷮2﷯=c −﷐1﷮2﷯ Shortest distance is D = ﷐﷮﷐ℎ﷮2﷯+﷐﷐𝑐−𝑘﷯﷮2﷯﷯ Putting ﷐ℎ﷮2﷯=𝑐−﷐1﷮2﷯ & 𝑘=𝑐−﷐1﷮2﷯ D = ﷐﷮﷐𝑐 −﷐1﷮2﷯﷯+﷐﷐𝑐−﷐𝑐−﷐1﷮2﷯﷯﷯﷮2﷯﷯ = ﷐﷮﷐𝑐−﷐1﷮2﷯﷯+﷐﷐﷐1﷮2﷯﷯﷮2﷯﷯ = ﷐﷮﷐𝑐 −﷐1﷮2﷯﷯+﷐1﷮4﷯﷯ = ﷐﷮𝑐 −﷐1﷮2﷯+﷐1﷮4﷯﷯ = ﷐﷮𝑐+﷐﷐− 2 + 1﷮4﷯﷯﷯ = ﷐﷮𝑐−﷐1﷮4﷯﷯ = ﷐﷮﷐4𝑐 − 1﷮4﷯﷯ Hence, shortest distance is ﷐﷮﷐𝟒𝒄 − 𝟏﷮𝟒﷯ ﷯

About the Author

CA Maninder Singh's photo - Expert in Practical Accounts, Taxation and Efiling
CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
Jail