The curves $x=y^2$ and $x y=a^3$ cut orthogonally at a point, then a =  

$\frac{1}{2}$

We have,

$x=y^2$     ...(i)     and,          $x y=a^3$        ...(ii)

These two curves intersect at $P\left(a, a^2\right)$

Now,

$x=y^2 \Rightarrow \frac{d y}{d x}=\frac{1}{2 y} \Rightarrow m_1=\left(\frac{d y}{d x}\right)_P=\frac{1}{2 a^2}$

and,

$x y=a^3 \Rightarrow x \frac{d y}{d x}+y=0 \Rightarrow \frac{d y}{d x}=-\frac{y}{x} \Rightarrow m_2=\left(\frac{d y}{d x}\right)_P=-a$

If the two curves intersect orthogonally, then

$m_1 m_2=-1 \Rightarrow \frac{1}{2 a^2} \times-a=-1 \Rightarrow a=\frac{1}{2}$