Tangent drawn at point P(1,3) of a parabola intersects its tangent at vertex at M(-1,5) and cuts the axis of parabola at T. If R(-5,5) is a point on SP; where S is focus of the parabola, then

Equation of tangent at P is x + y = 4
Clearly mirror image of R(-5,5) lies on line PQ.
Now mirror image $R^{\text{'}}$  of R

$\begin{array}{l}\Rightarrow \frac{{\alpha }^{\mathrm{\prime }}+5}{1}=\frac{{\beta }^{\mathrm{\prime }}-5}{1}=\frac{-2(-5+5-4)}{2}=4\\ \Rightarrow \left({\alpha }^{\mathrm{\prime }},{\beta }^{\mathrm{\prime }}\right)=(-1,9)\\ \text{ Let PM cuts the axis at }\mathrm{T};\text{ As }\mathrm{M}\text{ is midpoint of PT }\end{array}$

$\begin{array}{l}\text{ We know that }\mathrm{SP}=\mathrm{ST}\text{ and }\mathrm{\angle }SMP=\frac{\pi }{2}\\ \text{ Equation SP }=y-3=-\frac{1}{3}(x-1)\\ \Rightarrow x+3y-10=0\\ \text{ Let }S=(10-3\beta ,\beta )\\ \text{ Again }TS\parallel PQ\Rightarrow \frac{\beta -7}{13-3\beta }=\frac{9-3}{-1-1}=-3\\ \text{ focus is }(-2,4)\end{array}$