Q. 264.7( 3 Votes )

Find the area bou

Answer :

The given curves are,



2y+3=x … (2)


Solving (1) and (2) we get,


2y +3 = y2


y2 – 2y – 3 =0


y2 – 3y + y – 3 =0


y(y – 3) + 1(y – 3) = 0


(y + 1)(y – 3) = 0


y = –1, 3


As y >0


So, y = 3


Substituting y = 3 in (2) we get,


x = 2(3) + 3


x = 9


So, (1) and (2) intersect at (9, 3)






= (9+9) – 9 = 9 sq. units


So, required area = 9 sq. units.


OR


The given curves are,


x2 + y2 = 8 … (1)


x2 = 2y … (2)


Solving (1) and (2) we get,


2y + y2 – 8 =


y2 + 2y – 8 = 0


y2 + 4y – 2y – 8 = 0


y(y + 4) – 2(y + 4) = 0


(y + 4)(y – 2) = 0


y = – 4, 2


But y > 0 hence, y = 2


Put y = 2 in (2) we get,


x2 = 4


x = + 2, – 2


So, the point of intersections are (-2, 2) and (2, 2).



So, required area =




As,


So,







Hence,


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