Draw a rough sketch of the region {(x,y) : y^{2} ≤ 5x, 5x^{2} + 5y^{2} ≤ 36} and find the area enclosed by the region using the method of integration.

To find the area enclosed by the region{(x,y) : y^{2} ≤ 5x, 5x^{2} + 5y^{2} ≤ 36}

The given equations are,

y^{2} = 5x ...(i)

And 5x^{2} + 5y^{2} = 36 ...(ii)

Substituting the value of y^{2} from (i) into (ii)

5x^{2} + 25x = 36

5x^{2} + 25x – 36 = 0

x =

Equation (i) represents a parabola with vertex (0, 0) and axis as x - axis.

Equation (ii) represents a circle with centre (0, 0) and radius 6/√5 and meets axes at and . X ordinate of the point of intersection of circle and parabola is A where a = .

A rough sketch of curves is: -

Required area = Region OCBAO

= 2 (Region OBAO)

= 2 (Region ODAO + Region DBAD)

The area enclosed by the region is sq. Units

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