Given the boundaries of the area to be found are,

• the first parabola, y² = 4x ---(1)

• the second parabola, x² = 4y ---- (2)

From the equation, of the first parabola, y² = 4x

• the vertex at (0,0) i.e. the origin

• Symmetric about the x-axis, as it has the even power of y.

From the equation, of the second parabola, x² = 4y

• the vertex at (0,0) i.e. the origin

• Symmetric about the y-axis, as it has the even power of x.

Now to find the point of intersection of (1) and (2), substitute in (1)

x⁴ = 64x

x(x³- 64) = 0

x = 0 (or) x = 4

Substituting x in (2), we get y = 0 (or) y = 4

So the two points, A and B where (1) and (2) meet are A = (4,4) and O= (0,0)

Consider the first parabola, y² = 4x, can be re-written as

----- (3)

Consider the parabola, x² = 4y, can be re-written as

----- (4)

Let us drop a perpendicular from A on to x-axis. The base of the perpendicular is D = (4, 0)

Now, the area to be found will be the area is

Area of the required region = Area of OBACO.

Area of OBACO= Area of OBADO- Area of OCADO

Area of OBACOis

[Using the formula, ]

The Required Area of OBACOsq. units