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Question

Prove that the curves y2=4x and x2=4y divide the area of the square bounded by the sides x=0,x=4,y=4 and y=0 into three equal parts.

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Solution

y2=4x (1) and x2=4y (2)
Solving (1) & (2),
y4=16x2y4=16×(4y)y=4 and y=0x=4 and x=0
Therefore, the intersection points of the curves are (0,0) and (4,4).We have to show that
ar (ABCEA)=ar (AECFA)=ar (ADCFA)=13 ar (ABCDA)ar (ABCEA)=404 dx404x dx=4[x]40[2×23(x)32]40=16323=163 sq.units (1)

ar (AECFA)=40(4xx24) dx=323[x312]40=323163=163 sq.units (2)

ar ADCFA=40x24 dx=[x312]40=163 sq.units (3)

From (1),(2) & (3), we conclude that
ar (ABCEA)=ar (AECFA)=ar (ADCFA) (4)

Also, ar (ABCDA)=4×4=16 sq.units
and ar (ABCDA)=ar (ABCEA)+ar (AECFA)+ar (ADCFA) (5)

From (4) & (5), we get
ar (ABCEA)=ar (AECFA)=ar (ADCFA)=13 ar (ABCDA)

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