The area enclosed between the curves y = ax ^ 2 and x = ay ^ 2 (a >...

As from the figure, area enclosed between the curves is $OABCO$.

Thus, the points of intersection of

$\text{y} = {\text{ax}}^{2} and$\text{x} = {\text{ay}}^{2}$,<br />eliminating$y$, we get<br /><br />$x = a{( {ax}^{2} )}^{2}$<br /><br />$⇒ x - {a}^{3}{x}^{4} = 0$<br /><br />$⇒ x( 1 - {a}^{3}{x}^{3} ) = 6$<br /><br />$⇒ x = 0\text{and}x = \dfrac{1}{a}$<br /><br /><img alt="" src="https://cdn-question-pool.Quantrex.app/nta_abhyas/jee_main/images/math-3.png" /><br /><br />$∴$Points of intersection are$( 0, 0 )$and \left(\dfrac{1}{\text{a}} \text{,} \dfrac{1}{\text{a}}\right)$.

Thus, required area $OABCO$ = Area of curve $OCBDO$ - area of curve $OABDO$

$⇒$ ${∫}_{0}^{1 / \text{a}} \left(\sqrt{\dfrac{\text{x}}{\text{a}}} - {\text{ax}}^{2}\right) \text{dx} = 1$ (given)

$⇒$ ${\left[\dfrac{1}{\sqrt{\text{a}} \cdot} \cdot \dfrac{{\text{x}}^{3 / 2}}{3 / 2} - \dfrac{{\text{ax}}^{3}}{3}\right]}_{0}^{1 / \text{a}} = 1$

$⇒$ $\dfrac{2}{3 {\text{a}}^{2}} - \dfrac{1}{3 {\text{a}}^{2}} = 1$

$⇒ {\text{a}}^{2} = \dfrac{1}{3} ⇒ \text{a} = \dfrac{1}{\sqrt{3}} ( ∵ \text{a} > 0 )$

$\dfrac{1}{\sqrt{3}} = \dfrac{3\sqrt{3}}{\lambda}$

$\lambda = 9$