$\begin{aligned} & y^2=6 x \\ & \Rightarrow 2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=6 \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3}{y} \\ & \text { Also, } 9 x^2+\mathrm{b} y^2=16 \\ & \Rightarrow 18 x+2 \mathrm{~b} y \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-9 x}{\mathrm{~b} y}\end{aligned}$
As given curves intersect each other at right angle, their tangents also intersect at right angles.
$$
\begin{array}{ll}
& \frac{3}{y} \times \frac{-9 x}{\mathrm{~b} y}=-1 \\
\Rightarrow & \mathrm{b} y^2=27 x \\
\therefore \quad & (\mathrm{i}) \Rightarrow \mathrm{b}(6 x)=27 x \\
& \Rightarrow \mathrm{b}=\frac{9}{2}
\end{array}
$$