Area Question 25
Question 25
- The area enclosed between the curves $y=a x^{2}$ and $x=a y^{2}(a>0)$ is 1 sq unit. Then, the value of $a$ is (a) $\frac{1}{\sqrt{3}}$ (b) $\frac{1}{2}$ (c) 1 (d) $\frac{1}{3}$
(2004, 1M)
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Solution:
- As from the figure, area enclosed between the curves is $O A B C O$.
Thus, the point of intersection of
$$ y=a x^{2} \text { and } x=a y^{2} $$
$$ \begin{array}{ll} \Rightarrow & x=a\left(a x^{2}\right)^{2} \ \Rightarrow & x=0, \frac{1}{a} \Rightarrow y=0, \frac{1}{a} \end{array} $$
So, the points of intersection are $(0,0)$ and $\frac{1}{a}, \frac{1}{a}$.
$\therefore$ Required area $O A B C O=$ Area of curve $O C B D O$
- Area of curve $O A B D O$
$$ \begin{aligned} \Rightarrow & \int_{0}^{1 / a} \sqrt{\frac{x}{a}}-a x^{2} d x & =1 \ \Rightarrow & \frac{1}{\sqrt{a}} \cdot \frac{x^{3 / 2}}{3 / 2}-\frac{a x^{3}}{3}{ }^{1 / a} & =1 \ \Rightarrow & \frac{2}{3 a^{2}}-\frac{1}{3 a^{2}} & =1 \ & a^{2} & =\frac{1}{3} \Rightarrow \quad a=\frac{1}{\sqrt{3}} \quad[\because a>0] \end{aligned} $$
