# Misc 6 - Chapter 8 Class 12 Application of Integrals

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

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 y2=4𝑎𝑥 𝑚𝑥2=4𝑎𝑥 𝑚2 𝑥2=4𝑎𝑥 𝑚2 𝑥2−4𝑎𝑥=0 𝑥 𝑚2𝑥−4𝑎=0 i.e. 𝑥=0 , 𝑚2𝑥−4𝑎=0 𝑚2𝑥=4𝑎 𝑥= 4𝑎 𝑚2 Putting values of 𝑥 in 𝑦=𝑚𝑥 𝑦=𝑚 ×0=0 𝑦=𝑚 × 4𝑎 𝑚2= 4𝑎𝑚 So, the intersecting points are O 0 , 0 and A 4𝑎 𝑚2 , 4𝑎𝑚 Step 3: Finding the Area Area OBAD Area OBAD = 0 4𝑎 𝑚2𝑦 𝑑𝑥 y → Equation of parabola 𝑦2 = 4x 𝑦 = 4𝑎𝑥 Therefore, Area OBAD = 0 4𝑎 𝑚2 4𝑎𝑥 𝑑𝑥 = 0 4𝑎 𝑚2 4𝑎 . 𝑥 𝑑𝑥 = 4𝑎 0 4𝑎 𝑚2 𝑥 𝑑𝑥 = 4𝑎 𝑥 12+1 12+10 4𝑎 𝑚2 = 4𝑎 𝑥 32 320 4𝑎 𝑚2 = 4𝑎 × 23 𝑥 320 4𝑎 𝑚2 = 2(2 𝑎)3 4𝑎 𝑚2 32− 0 32 = 4 𝑎3 4𝑎 𝑚2 4𝑎 𝑚2−0 = 4 𝑎3 4𝑎 𝑚2 × 2 𝑎𝑚 = 32 𝑎 . 𝑎 . 𝑎3 𝑚3 = 32 𝑎23 𝑚3 Area OAD Area OAD = 0 4𝑎 𝑚2𝑦 𝑑𝑥 y → Equation of line y = mx Therefore, Area OAD = 0 4𝑎 𝑚2m𝑥 𝑑𝑥 = m 0 4𝑎 𝑚2𝑥 𝑑𝑥 = 𝑚 𝑥220 4𝑎 𝑚2 = 𝑚2 4𝑎 𝑚22− 02 = 𝑚2 4𝑎2 𝑚4 = 8 𝑎2 𝑚3 Thus, Area Required = Area OBAD – Area OAD = 32 𝑎23 𝑚3 – 8 𝑎2 𝑚3 = 32 − 24 𝑎23 𝑚3 = 8 𝑎2 𝑚3

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.