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

Let curves intersect each other at $P\left(x_1,y_1\right)$. P lies on both the curves $y_1^2=6x_1\to \left(1\right)$

$$\begin{gathered} 9x_1^2+b{y}_1^2=16\to \left(2\right) \\ \text{ For }{1}^{\text{st }}\text{ curve }{\left(\frac{dy}{dx}\right)}_P=\frac{3}{y_1}=m_1 \\ \text{For }{2}^{\text{nd }}\text{ curve }{\left(\frac{dy}{dx}\right)}_P=\frac{-9x_1}{b{y}_1}=m_2 \end{gathered}$$$m_1{m}_2=-1\Rightarrow \frac{-27x_1}{b{y}_1^2}=-1\Rightarrow b=27\cdot \frac{x_1}{y_1^2}=27\cdot \frac{1}{6}=\frac{9}{2}=4.5$