# Ex 8.2 , 7

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 8.2 , 7 Area lying between the curves 𝑦2 = 4𝑥 and 𝑦=2𝑥 is (A) 𝟐𝟑 (B) 𝟏𝟑 (C) 𝟏𝟒 (D) 𝟑𝟒 Step 1: Drawing figure Parabola is 𝑦2 =4x Also, 𝑦=2𝑥 passes through (0, 0) & (1, 2) Point (1, 2) lies in parabola y2 = 4x Hence, intersecting point A = (1, 2) Area required Area required = Area OBAD – Area OAD Area OBAD Area OBAD = 01𝑦 𝑑𝑥 y → Equation of parabola 𝑦2 = 4x 𝑦 = 4x 𝑦 = 2 x Therefore, Area OBAD = 012 x 𝑑𝑥 = 2 01 𝑥 12 𝑑𝑥 = 2 𝑥 12+1 12+101 = 2 𝑥 32 3201 = 2 × 23 𝑥 3201 = 43 1 32− 0 32 = 43 Area OAD Area OAD = 01𝑦 𝑑𝑥 y → Equation of line y = 2x Therefore, Area OAD = 012𝑥 𝑑𝑥 = 2 01𝑥 𝑑𝑥 = 2 𝑥2201 = 2 × 12 𝑥201 = 12− 02 = 1 Area required = Area OBAD − Area OAD = 43 − 1 = 13 So, B is correct option

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