First of all the given eqn.  2{(x-6 )2 + (y-6)2}=(x+y-4)2 is not a defined parabola as it consists both x and y variables with power 2 and neither it gets cancelled out from R.H.S
so please first check the question.
and mention again the suitable parsbolic eqn.
only one varable either x or y should have power 2.
						

							
Point 1,3 lie on the directrix of the given parabola and focus is 6,6  we have to find angle msp now think about normal parabola angle msp is 90° and this will hold for all parabola and the given equation represent a parabola of focus 6,6 and directrix x+y-4
						

							
the angle msp is independent of coordinate system hence we shift the parabola
2{(x-6)^2+(y-6)^2}=(x+y-4)^2 parallel to y axis the angle msp never change hence the equation of parabola becomes y^2=4ax
point p(at^2,2at)
equation of tangent to the parabola y^2=4ax is
ty=x+at^2  ---(1)
dirctrix cut axis at(-a,0)
putting x in point(-a,0) in eqn(1) we get coordinates of m
x=-a
y=a(t^2-1)/t
slope msp= (2at-0)/(at^2-a)=(2t)/(t^2-1)
slope msm= -(t^2-1)/2t
msp×msm= -1
hence angle is 90°