Mathematics - Area Under Curves Question with Solution | TestHub

We have given, $y=a{x}^2$ ....(i) and $x=a{y}^2$ ......(ii) Put the value of $y$ by Eq. (i) in Eq. (ii), we get$x=a\times a^2{x}^4\Rightarrow x^4{a}^3-x=0$ $x\left(x^3{a}^3-1\right)=0\Rightarrow x=0,\frac{1}{2}$ When, $x=0\Rightarrow y=0$ and $x=\frac{1}{a}\Rightarrow y=\frac{1}{a}$ Here, points of intersection of curves $y=a{x}^2$ and $x=a{y}^2$ are $(0,0)$ and $\left(\frac{1}{a},\frac{1}{a}\right)$. $\therefore$ Required area $A={\int }_{x=a}^{x=b}\left[f_2(x)-f_1(x)\right]dx$ $3={\int }_0^{1/a}\left(\frac{\sqrt{x}}{\sqrt{a}}-a{x}^2\right)dx$ $3={\left[\frac{1}{\sqrt{a}}\times \frac{2}{3}{x}^{3/2}-\frac{a{x}^3}{3}\right]}_0^{1/a}$ $3=\frac{2}{3\sqrt{a}}\left[{\left(\frac{1}{a}\right)}^{3/2}\right]-\frac{a}{3}\left[{\left(\frac{1}{a}\right)}^3\right]$ $3=\frac{2}{3\sqrt{a}}\times \frac{1}{a\sqrt{a}}-\frac{a}{3}\times \frac{1}{a^3}$ $3=\frac{2}{3a^2}-\frac{1}{3a^2}\Rightarrow 3=\frac{2-1}{3a^2}\Rightarrow 9a^2=1$ $a^2=\frac{1}{9}\Rightarrow a=\frac{1}{3}$