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

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  1. Class 12
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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﷯ ﷯

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.