Question 34 (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at April 16, 2024 by Teachoo

Find the area of the region bounded by the curves ๐ฅ
^{
2
}
+ ๐ฆ
^{
2
}
= 4, y = √3๐ฅ ๐๐๐ ๐ฅ − ๐๐ฅ๐๐ ๐๐ ๐กℎ๐ ๐๐๐๐ ๐ก ๐๐ข๐๐๐๐๐๐ก

Question 34 (Choice 1) Find the area of the region bounded by the curves ๐ฅ^2+๐ฆ^2=4, ๐ฆ=โ3 ๐ฅ ๐๐๐ ๐ฅ โ ๐๐ฅ๐๐ ๐๐ ๐กโ๐ ๐๐๐๐ ๐ก ๐๐ข๐๐๐๐๐๐ก
Given Equation of Circle
๐ฅ2+๐ฆ2=4
๐ฅ2+๐ฆ2=2^2
So, Radius = 2
โด Point A (2, 0) and B is (0, 2)
Let point where line and circle intersect be point M
Required Area = Area of shaded region
= Area OMA
Finding Point M
Point M is intersection of line and circle
Putting ๐=โ๐ ๐ in Equation of Circle
๐ฅ2+๐ฆ2=4
๐ฅ2+(โ3 ๐ฅ)2=4
4๐ฅ2=4
๐ฅ2=1
โด ๐ฅ=ยฑ1
When ๐=๐
๐ฆ=โ3 ๐ฅ=โ3
So, point is (1, โ3)
When ๐=โ๐
๐ฆ=โ3 ๐ฅ=โโ3
So, point is (โ1, โโ3)
As Point M is in 1st Quadrant
โด M = (1, โ3)
โด Required Area = Area OMP + Area PMA
= โซ1_0^1โ๐ฆ1๐๐ฅ+โซ1_1^2โ๐ฆ2๐๐ฅ
Finding y1 and y2
For y1 โ Equation of Line
y = โ3x
So, y1 = โ3x
For y2 โ Equation of Circle
๐ฅ^2+๐ฆ^2=4
๐ฆ^2=4โ๐ฅ^2
โด ๐ฆ=ยฑโ(2^2โ๐ฅ^2 )
As Required Area is in first Quadrant
โด ๐ฆ2=โ(2^2โ๐ฅ^2 )
Thus,
Required Area = โซ1_0^1โใโ3 ๐ฅ ๐๐ฅใ+โซ1_1^2โใโ(2^2โ๐ฅ^2 ) ๐๐ฅใ
=โ3 โซ1_0^1โใ๐ฅ ๐๐ฅใ+โซ1_1^2โใโ(2^2โ๐ฅ^2 ) ๐๐ฅใ
=โ3 [๐ฅ^2/2]_0^1+โซ1_1^2โใโ(2^2โ๐ฅ^2 ) ๐๐ฅใ
=โ3 [1^2/2โ0^2/2]_0^1+โซ1_1^2โใโ(2^2โ๐ฅ^2 ) ๐๐ฅใ
=โ3/2+โซ1_1^2โใโ(2^2โ๐ฅ^2 ) ๐๐ฅใ
It is of form
โซ1โใโ(๐^2โ๐ฅ^2 ) ๐๐ฅ=1/2 ๐ฅโ(๐^2โ๐ฅ^2 )ใ+๐^2/2 ใ๐ ๐๐ใ^(โ1)โกใ๐ฅ/๐+๐ใ
Replacing a by 2 , we get
ใ=โ3/2+[๐ฅ/2 โ((2)^2โ๐ฅ^2 )+(2)^2/2 sin^(โ1)โกใ๐ฅ/2ใ ]ใ_1^2
=โ3/2+2/2 โ((2)^2โ(2)^2 )+2 sin^(โ1)โกใ2/2ใโ1/2 โ((2)^2โ("1" )^2 )โ2 sin^(โ1)โกใ1/2ใ
=โ3/2+0+2 sin^(โ1)โกใ(1)โ1/2 โ(4โ1)โใ 2 sin^(โ1)โก(1/2)
=โ3/2+2 ร๐/2โโ3/2โ2ร๐/6
=๐ โ๐/3
=๐๐ /๐ square units

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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