Consider a parabola  $y^2=\sqrt{2}x$. From a point where its directrix intersects the line  $2\sqrt{2}x-\sqrt{2}y+3=0$, a normal to the parabola is drawn, meeting the parabola at M (foot of normal) and N. R is point of intersection of tangent at M and chord ON produced (O is origin). If angle $\angle MRN$ is  $\theta$ then value of $\left|\tan \theta \right|$ is

$y^2=\sqrt{2}x\Rightarrow a=\frac{1}{2\sqrt{2}}$

Its directrix $x=-\frac{1}{2\sqrt{2}}$ intersects
The line $2\sqrt{2}x-\sqrt{2}y+3=0$ at  $P(-\frac{1}{2\sqrt{2}},\sqrt{2})$

Equation of normal is $y=mx-2am-a{m}^3$ passing through  $(-\frac{1}{2\sqrt{2}},\sqrt{2})$

$$\begin{gathered} \Rightarrow \sqrt{2}=\frac{-1}{2\sqrt{2}}m-2\frac{1}{2\sqrt{2}}m-\frac{1}{2\sqrt{2}}{m}^3 \\ {m}^3+3m+4=0 \\ \Rightarrow m=-1 \end{gathered}$$

Foot of normal is  $M(a{m}^2,-2am)\equiv M(\frac{1}{2\sqrt{2}},\frac{1}{\sqrt{2}})$

$\because t_1=1\Rightarrow t_2=-t_1-\frac{2}{t_1}=-3$

$\Rightarrow$ Other end of normal is  $N(\frac{9}{2\sqrt{2}},-\frac{3}{\sqrt{2}})$
Slope of  $RN=m_{ON}=-\frac{2}{3}$
Slope of $RM = 1$
$\Rightarrow \tan \theta =\left|\frac{1+\frac{2}{3}}{1-\frac{2}{3}}\right|=5$