If the curves \(y^2=6 x, 9 x^2+b y^2=16\) intersect each other at right angles, then the value of \(b\) is

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