If the curves $y^2=6x$  and  $9x^2+b{y}^2=16$  intersect each other at right angles, then the value of b is 

$$\begin{gathered} y^2=6x,9x^2+b{y}^2=16 \\ 2y\frac{dy}{dx}=6 18x+2by\frac{dy}{dx}=0 \\ \frac{dy}{dx}=\frac{3}{y_1} 2by\frac{dy}{dx}=-18x \\ {m}_1=\frac{3}{y_1} \frac{dy}{dx}=\frac{-9x_1}{b{y}_1} \\ m_2=\frac{-9x_1}{b{y}_1} \\ {m}_1{m}_2=-1 \\ \frac{+27x_1}{b{y}_1^2}=+1\Rightarrow 27x_1=b.6x_1 \\ b=\frac{9}{2} \end{gathered}$$