The provided equations for two curves are \(y^2 = 6x\) (1) and \(9x^2 + by^2 = 16\) (2). To determine the value of \(b\), both equations are differentiated with respect to \(x\). Differentiation of equation (1) yields \(2y \frac{dy}{dx} = 6\), which simplifies to \(\frac{dy}{dx} = \frac{3}{y}\). Differentiation of equation (2) yields \(18x + 2by \frac{dy}{dx} = 0\), simplifying to \(\frac{dy}{dx} = -\frac{9x}{by}\). For the curves to intersect perpendicularly, the product of their tangent slopes at the intersection point must be \(-1\). Therefore, \(\frac{3}{y} \cdot \left(-\frac{9x}{by}\right) = -1\). This simplifies to \(-\frac{27x}{by^2} = -1\), or \(\frac{27x}{by^2} = 1\). Substituting \(y^2 = 6x\) from equation (1) into this equation gives \(\frac{27x}{b \cdot 6x} = 1\), which reduces to \(\frac{27}{6b} = 1\). Solving for \(b\) yields \(b = \frac{27}{6} = \frac{9}{2}\). The value of \(b\) is \(\frac{9}{2}\).