Use Reflection Properties Of Parabola !

Here Note That Given two Tangent's Lines are perpendicular $\displaystyle\because \quad ({ m }_{ 1 }{ m }_{ 2 }=\quad -1)$ .

So It is Property of Parabola That Two perpendicular tangent's must be intersect the directrix of parabola (Definition of Director Circle of Parabola) So directrix must be Pass through (0,1) .

Again By using Reflection Property of Parabola that is Image of Focus Lies in given Tangent Line Lies on directrix So image of Focus is Q(4,6) By taking image in Line $L_{ 1 }$ .

[See Below How I directly Compute This? I have Short Trick For it]

Eq. of Directrix : from two point's $(0,1)$ & $(4,6)$

$\displaystyle5x-4y+4\quad =\quad 0\\$ .

Now By Using Focal Directrix Property of Parabola i.e Distance of any point P(x,y) on parabola from Focus is equal to the perpendicular distance of Point P(x,y) from Directrix of parabola :

$\displaystyle\\ SP\quad =\quad PM\quad \\ \\ \boxed { \left| \cfrac { 5x-4y+4 }{ \sqrt { 41 } } \right| =\sqrt { { (x-5) }^{ 2 }+{ (x-5) }^{ 2 } } }$ .

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Short Trick :

For Calculating Image of a Point $P(a,b)$ in a given Line L : $y\quad =\quad x\quad +\quad c$ . is given By : $\boxed { \quad Q(b-c,\quad a+c)\quad }$ .

(You can Verify it by taking standard case C=0) !

this question can also be solved without reflection properties Since the directrix pass through (0,1) and perpendicular to the line joining the points { (0,1) , (5,5)} .Hence we can write the equation of directrix and using the Basic defination of parabola that "distance of pt. from fix Pt. is equal to distance from directrix.Hence A+B+C+D+E=59