Find the area of the region {(x,y):y2≤4x,4x2+4y2=9.

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Filo · Accepted Answer

Given curves are y2=4x     ....(1)4x2+4y2=9     ...(2)⇒x2+y2=49​Center of circle is (0,0) and radius of circle is 23​Put the value from eqn (1) in eqn (2),4x2+16x−9=0⇒x=21​,−29​But x=−29​, not possible .So, x=21​⇒y=±2​So, the points of intersection of both the curves are (21​,2​) and (21​,−2​).The shaded region OABCO represents the area bounded by the curves {(x,y):y2≤4x,4x2+4y2≤9}Since, the area OABCO is symmetrical about x-axis.∴Area OABCO=2×Area OBCArea OBCO=Area OMC+Area MBC=∫021​​2x​dx+∫21​23​​21​9−4x2​dx=2[3/2x3/2​ ]01/2​+∫1/23/2​(23​)2−(x)2​ dx=34​22​ 1​+⎣⎡​2x(23​)2−(x)2​ ​+249​ ​sin−1(23​ x​ )  ⎦⎤​1/23/2​=32​ ​+[0+89​sin−1(1)−221​2​ ​−89​sin−1(31​)]=32​ ​+169π ​−42​ ​−89​sin−1(31​)Required Area =2[32​ ​+169π ​−42​ ​−89​sin−1(31​)]=62​ ​+89π ​−49​sin−1(31​) sq.units

Question Text	Find the area of the region ${(x, y) : y^{2} \leq 4 x, 4 x^{2} + 4 y^{2} = 9.$
Answer Type	Text solution:1
Upvotes	150

Find the area of the region ${(x, y) : y^{2} \leq 4 x, 4 x^{2} + 4 y^{2} = 9.$

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