Question

Prove that the curves $y^2=4x\text{and}{x}^2=4y$ divide the area of the square bounded by the sides $x=0,x=4,y=4\text{and}y=0$ into three equal parts.

Answer 1

$y^2=4x\cdots \left(1\right)\text{and}{x}^2=4y\cdots \left(2\right)$
Solving $\left(1\right)\&\left(2\right),$
$y^4=16x^2\Rightarrow y^4=16\times \left(4y\right)\Rightarrow y=4\text{and}y=0\Rightarrow x=4\text{and}x=0$
Therefore, the intersection points of the curves are $(0,0)\text{and}(4,4).$We have to show that
$\text{ar}\left(ABCEA\right)=\text{ar}\left(AECFA\right)=\text{ar}\left(ADCFA\right)=\frac{1}{3}\text{ar}\left(ABCDA\right)\text{ar}\left(ABCEA\right)={\int }_0^{4}4dx-{\int }_0^4\sqrt{4x}dx=4{\left[x\right]}_0^4-{[2\times \frac{2}{3}(x{)}^{\frac{3}{2}}]}_0^4=16-\frac{32}{3}=\frac{16}{3}\text{sq.units}\cdots \left(1\right)$

$\text{ar}\left(AECFA\right)={\int }_0^4(\sqrt{4x}-\frac{x^2}{4})dx=\frac{32}{3}-{\left[\frac{x^3}{12}\right]}_0^4=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}\text{sq.units}\cdots \left(2\right)$

$\text{ar}ADCFA={\int }_0^4\frac{x^2}{4}dx={\left[\frac{x^3}{12}\right]}_0^4=\frac{16}{3}\text{sq.units}\cdots \left(3\right)$

From $\left(1\right),\left(2\right)\&\left(3\right),$ we conclude that
$\text{ar}\left(ABCEA\right)=\text{ar}\left(AECFA\right)=\text{ar}\left(ADCFA\right)\cdots \left(4\right)$

$\text{Also, ar}\left(ABCDA\right)=4\times 4=16\text{sq.units}$
$\text{and ar}\left(ABCDA\right)=\text{ar}\left(ABCEA\right)+\text{ar}\left(AECFA\right)+\text{ar}\left(ADCFA\right)\cdots \left(5\right)$

From $\left(4\right)\&\left(5\right),$ we get
$\text{ar}\left(ABCEA\right)=\text{ar}\left(AECFA\right)=\text{ar}\left(ADCFA\right)=\frac{1}{3}\text{ar}\left(ABCDA\right)$