Two tangents on a parabola are $\mathrm{x}-\mathrm{y}-1=0,\mathrm{x}+\mathrm{y}+11=0$ if $\mathrm{S}(1,3)$ is focus of the parabola. P, Q are ends of focal chord then 

Use the property foot of the perpendicular from focus to tangent lies on tangent at vertex find foot of perpendicular from $\mathrm{S}(1,3)$ to $\mathrm{x}-\mathrm{y}-1=0$. 

Say A=$\left(\frac{5}{2},\frac{3}{2}\right)$  and the foot of perpendicular from $\mathrm{S}(1,3)$ to $\mathrm{x}+\mathrm{y}+11=0$ 
say B $\left(\frac{-13}{2},\frac{-9}{2}\right)$.
then AB is tangent at vertex X.
Equation of AB is $3y-2x+\frac{1}{2}=0$ 
$$\begin{gathered} a= perpendicular dis\tan ce for S to AB \\ =\frac{15}{2\sqrt{13}} \\ \Rightarrow L.L.R=4a=\frac{30}{\sqrt{13}} \\ now \frac{1}{SP}+\frac{1}{SQ}=\frac{1}{a}=\frac{2\sqrt{13}}{15} \end{gathered}$$
Equation of Latusrecta=$3y-2x-7=0$
Equation of directrix is
 $$\begin{gathered} 3y-2x+2\left(\frac{1}{2}\right)-\left(-7\right)=0 \\ \Rightarrow 3y-2x+8=0 \end{gathered}$$