Parabola Part 1

$\tan \frac{\pi}{4} = \left|\frac{m-3}{1+3m}\right|$$\Rightarrow \;\frac{m-3}{1+3m} = \pm 1$$\Rightarrow \;\frac{m-3}{1+3m}=+1 \;\text{ and }\; \frac{m-3}{1+3m}=-1$$\Rightarrow m-3=1+3m \;\text{ and }\; m-3=-1-3m$$\Rightarrow 2m=-4 \;\text{ and }\; 4m=2$$m=-2 \;\text{ and }\; m=\frac{1}{2}$We know, Equation of tangent to the parabola $y^2=4m$ is $y=mx+\frac{a}{m}$ and point of contact is $\left(\frac{a}{m^2},\frac{2a}{m}\right)$$\therefore$ Equation of tangent$y=-2x-\frac{a}{2}$and $y=\frac{x}{2}+2a$$\therefore$ Point of contact A and B are$A\left(\frac{a}{(-2)^2},\frac{2a}{-2}\right)=A\left(\frac{a}{4},-a\right)$$B\left(\frac{a}{\left(\frac{1}{2}\right)^2},\frac{2a}{\frac{1}{2}}\right)=B(4a,4a)$As points A, B and S are colinear so area of triangle formed by those 3 points are zero. $\;\text{ Area of }\; \triangle ABS = \frac{1}{2} \left| \begin{vmatrix} \frac{a}{4} & -a & 1 \\ 4a & 4a & 1 \\ a & 0 & 1 \end{vmatrix} \right|$$=\frac{a}{4}(4a-0)+a(4a-a)+1(0-4a^2)$$=a^2+3a^2-4a^2=0$$\therefore$ Area of triangle is independent of value of a.So, for all value of $a>0$ (already given a must be greater than 0 ) point $A, B$ and $S$ will be collinear.