Misc 6 - Find area enclosed between parabola y2 = 4ax and y=mx - Area between curve and line

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  1. Chapter 8 Class 12 Application of Integrals
  2. Serial order wise
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Misc 6 Find the area enclosed between the parabola 𝑦2=4𝑎𝑥 and the line 𝑦=𝑚𝑥 Step 1: Draw the Figure Parabola is 𝑦2 =4ax Also, 𝑦=𝑚𝑥 is a straight line Step 2: Finding point of intersection A Point A is intersection of line and parabola Putting y = mx in equation of parabola y﷮2﷯=4𝑎𝑥 𝑚𝑥﷯﷮2﷯=4𝑎𝑥 𝑚﷮2﷯ 𝑥﷮2﷯=4𝑎𝑥 𝑚﷮2﷯ 𝑥﷮2﷯−4𝑎𝑥=0 𝑥 𝑚﷮2﷯𝑥−4𝑎﷯=0 i.e. 𝑥=0 , 𝑚﷮2﷯𝑥−4𝑎=0 𝑚﷮2﷯𝑥=4𝑎 𝑥= 4𝑎﷮ 𝑚﷮2﷯﷯ Putting values of 𝑥 in 𝑦=𝑚𝑥 𝑦=𝑚 ×0=0 𝑦=𝑚 × 4𝑎﷮ 𝑚﷮2﷯﷯= 4𝑎﷮𝑚﷯ So, the intersecting points are O 0 , 0﷯ and A 4𝑎﷮ 𝑚﷮2﷯﷯ , 4𝑎﷮𝑚﷯﷯ Step 3: Finding the Area Area OBAD Area OBAD = 0﷮ 4𝑎﷮ 𝑚﷮2﷯﷯﷮𝑦 𝑑𝑥 ﷯ y → Equation of parabola 𝑦﷮2﷯ = 4x 𝑦 = ﷮4𝑎𝑥﷯ Therefore, Area OBAD = 0﷮ 4𝑎﷮ 𝑚﷮2﷯﷯﷮ ﷮4𝑎𝑥﷯ 𝑑𝑥 ﷯ = 0﷮ 4𝑎﷮ 𝑚﷮2﷯﷯﷮ ﷮4𝑎﷯ . ﷮𝑥﷯ 𝑑𝑥 ﷯ = ﷮4𝑎﷯ 0﷮ 4𝑎﷮ 𝑚﷮2﷯﷯﷮ ﷮𝑥﷯ 𝑑𝑥 ﷯ = ﷮4𝑎﷯ 𝑥﷮ 1﷮2﷯+1﷯﷮ 1﷮2﷯+1﷯﷯﷮0﷮ 4𝑎﷮ 𝑚﷮2﷯﷯﷯ = ﷮4𝑎﷯ 𝑥﷮ 3﷮2﷯﷯﷮ 3﷮2﷯﷯﷯﷮0﷮ 4𝑎﷮ 𝑚﷮2﷯﷯﷯ = ﷮4𝑎﷯ × 2﷮3﷯ 𝑥﷮ 3﷮2﷯﷯﷯﷮0﷮ 4𝑎﷮ 𝑚﷮2﷯﷯﷯ = 2(2 ﷮𝑎﷯)﷮3﷯ 4𝑎﷮ 𝑚﷮2﷯﷯﷯﷮ 3﷮2﷯﷯− 0﷯﷮ 3﷮2﷯﷯﷯ = 4 ﷮𝑎﷯﷮3﷯ 4𝑎﷮ 𝑚﷮2﷯﷯ ﷮ 4𝑎﷮ 𝑚﷮2﷯﷯﷯−0﷯ = 4 ﷮𝑎﷯﷮3﷯ 4𝑎﷮ 𝑚﷮2﷯﷯ × 2 ﷮𝑎﷯﷮𝑚﷯﷯ = 32 𝑎 . ﷮𝑎﷯ . ﷮𝑎﷯﷮3 𝑚﷮3﷯﷯ = 32 𝑎﷮2﷯﷮3 𝑚﷮3﷯﷯ Area OAD Area OAD = 0﷮ 4𝑎﷮ 𝑚﷮2﷯﷯﷮𝑦 𝑑𝑥﷯ y → Equation of line y = mx Therefore, Area OAD = 0﷮ 4𝑎﷮ 𝑚﷮2﷯﷯﷮m𝑥 𝑑𝑥﷯ = m 0﷮ 4𝑎﷮ 𝑚﷮2﷯﷯﷮𝑥 𝑑𝑥﷯ = 𝑚 𝑥﷮2﷯﷮2﷯﷯﷮0﷮ 4𝑎﷮ 𝑚﷮2﷯﷯﷯ = 𝑚﷮2﷯ 4𝑎﷮ 𝑚﷮2﷯﷯﷯﷮2﷯− 0﷮2﷯﷯ = 𝑚﷮2﷯ 4𝑎﷯﷮2﷯﷮ 𝑚﷮4﷯﷯ = 8 𝑎﷮2﷯﷮ 𝑚﷮3﷯﷯ Thus, Area Required = Area OBAD – Area OAD = 32 𝑎﷮2﷯﷮3 𝑚﷮3﷯﷯ – 8 𝑎﷮2﷯﷮ 𝑚﷮3﷯﷯ = 32 − 24﷯ 𝑎﷮2﷯﷮3 𝑚﷮3﷯﷯ = 8 𝑎﷮2﷯﷮ 𝑚﷮3﷯﷯

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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.