The angle subtended by the normal chord at the point (9, 9) on the parabola ${\mathrm{y}}^2=9\mathrm{x}$, at the focus of the parabola is

$$\begin{gathered} Given parabola y^2=9x \\ \Rightarrow a=\frac{9}{4} \\ focus \left(S\right)=(\frac{9}{4},0) \\ NowP \left({\mathrm{at}}^2, 2\mathrm{at}\right)=\left(9,9\right) \\ \Rightarrow 2\mathrm{at} = 9 \\ \Rightarrow 2\left(\frac{9}{4}\right)\mathrm{t}=9 \\ \Rightarrow t=2 \\ Now t_1=-t-\frac{2}{t} \\ \Rightarrow t_1=-3 \\ Now Q=(\frac{81}{4},\frac{-27}{2}) \\ Now slope of SP=\frac{9}{9-{\displaystyle \frac{9}{4}}}=\frac{4}{3} \\ and slope of SQ=\frac{\frac{-27}{2}}{{\displaystyle \frac{81}{4}}-\frac{9}{4}}=\frac{-3}{4} \\ slope of SP\times slope of SQ=-1 \\ \therefore \mathrm{angle} \mathrm{at} \mathrm{focus} = {90}^{\circ } \end{gathered}$$