Answer :

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


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