The focus and directrix of a parabola are (1, 2) and $\mathrm{x}+2\mathrm{y}+9=0$ then equation of Tangent at vertex is 

S=(1, 2) directrix $\mathrm{x}+2\mathrm{y}+9=0$

let Z be foot of perpendicular from focus to directrix

$$\begin{gathered} \frac{\mathrm{h}-1}{1}=\frac{\mathrm{k}-2}{2}=\frac{-\left(1+4+9\right)}{1+4} \\ \frac{\mathrm{h}-1}{1}=\frac{\mathrm{k}-2}{2}=\frac{-14}{5} \\ \mathrm{h}=\frac{-9}{5}, \mathrm{k}-2=\frac{-28}{5} \\ \Rightarrow \mathrm{k}=\raisebox{1ex}{$-18$}\!\left/ \!\raisebox{-1ex}{$5$}\right. \end{gathered}$$

$$\begin{gathered} Let Z=\left(\frac{-9}{5},\frac{-18}{5}\right),S=\left(1, 2\right) \\ vertex=mid point of \overline{SZ} \\ \mathrm{V}=\left(\frac{{\displaystyle \frac{-9}{5}}+1}{2},\frac{{\displaystyle \frac{-18}{5}}+2}{2}\right) \\ \mathrm{V}=\left(\frac{-4}{10},\frac{-8}{10}\right)=(\frac{-2}{5},\frac{-4}{5}) \end{gathered}$$

Equation of tangent at vertex parallel to directrix is

$$\begin{gathered} \mathrm{x}+2\mathrm{y}+\mathrm{k}=0 \\ \sin ce it passes through (\frac{-2}{5},\frac{-4}{5}) then \\ \frac{-2}{5}-\frac{8}{5}+\mathrm{k}=0 \\ \Rightarrow k=2 \\ required line is x+2y+2=0 \end{gathered}$$