Let  $L_1:x+y=0and{L}_2:x-y=0$ are tangent to a parabola whose focus is  $S(1,2)$ . If the length of latus-rectums of the parabola can be expressed as  $\frac{m}{\sqrt{n}}$ (m and n are co prime).  Then the value of   $m+n-4$ is ____

Feet of the perpendicular from focus upon any tangent to parabola lies on the tangent at  the vertex.
Equation of  $S{N}_1$  is  $x+y=k\Rightarrow 1$
$(1,2)$  lies on it  $\therefore 1\Rightarrow x+y=3$
Solve  $x+y=3andx=y\therefore N_1=(\frac{3}{2},\frac{3}{2})$
Equation of  $S{N}_2$  is  $x-y=t\Rightarrow 2$
$(1,2)$  lies on it  $\therefore 2\Rightarrow y-x=1$
Solve  $y-x=1andx+y=0\therefore N_2=(\frac{-1}{2},\frac{1}{2})$
$\therefore$  Equation of tangent at vertex = equation of  $N_1{N}_2$
 $=2x-4y+3=0$
L.L.R = 4(distance from S(1,2) to tangent at vertex)
L.L.R =   $4\times \frac{|2-8+3|}{\sqrt{4+16}}=\frac{4\times 3}{2\sqrt{5}}=\frac{6}{\sqrt{5}}$