# Find the area bou

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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