The area enclosed between the curves $y = a$ and $x^2 =ay^2 (a > 0)$ is 1 square unit, then the value of a, is

$\frac{1}{\sqrt{3}}$

We have,

$\int\limits_{0}^{1/a}(y_2-y_1)dx=1$

$⇒\int\limits_{0}^{1/a}\left\{\sqrt{\frac{x}{a}}-ax^2\right\}dx=1$

$⇒\left[\frac{2}{3\sqrt{a}}x^{3/2}-\frac{a}{3}x^3\right]_{0}^{1/a}=1⇒\frac{2}{3\sqrt{a}}×\frac{1}{a^{3/2}}-\frac{a}{3}×\frac{1}{a^3}=1$

$⇒\frac{2}{3a^2}-\frac{1}{3a^2}=1⇒\frac{1}{3a^2}=1⇒a^2=\frac{1}{3}⇒a=\frac{1}{\sqrt{3}}$  $[∵a>0]$