If $\mathrm{P}(\mathrm{h},\mathrm{k})$ be a point on the parabola  $\mathrm{x}=4{\mathrm{y}}^2$, which is nearest to the point $Q(0, 33)$, then the distance of P from the directrix of the parabola ${\text{y}}^2=4(\mathrm{x}+\mathrm{y})$ is equal to:

$\begin{array}{l}{\mathrm{y}}^{\mathrm{2}}\mathrm{=}\frac{\mathrm{x}}{\mathrm{4}}\mathrm{ }\\ 4\mathrm{a}=\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow \mathrm{a}=\frac{\mathrm{1}}{\mathrm{16}}\\ \mathrm{normal} \mathrm{at} \left(\frac{\mathrm{1}}{\mathrm{16}}{\mathrm{t}}^{\mathrm{2}}\mathrm{,}\frac{\mathrm{1}}{\mathrm{8}}\mathrm{t}\right)\\ (0,33)\in \mathrm{y}+\mathrm{xt}=\frac{{\mathrm{t}}^{\mathrm{3}}}{\mathrm{16}}\mathrm{+}\frac{\mathrm{t}}{\mathrm{8}}\\ {\mathrm{t}}^{\mathrm{3}}+2\mathrm{t}=528\end{array}$
$\begin{array}{r}(\mathrm{t}-8)\left({\mathrm{t}}^2+8\mathrm{t}+66\right)=0\\ \mathrm{t}=8\end{array}$

t=8 is the only Solution
$\begin{array}{l}\therefore \mathrm{P}=\left(\frac{\mathrm{64}}{\mathrm{16}}\mathrm{,}\frac{\mathrm{8}}{\mathrm{8}}\right)\mathrm{=}(4,1)\\ \mathrm{Directrix} \mathrm{of} {(\mathrm{y}-2)}^2=4(\mathrm{x}+1)\mathrm{is}\mathrm{ }\mathrm{x}+1=-1\Rightarrow \mathrm{x}+2=0\\ \mathrm{required} \mathrm{distance}=6\end{array}$