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$$\tan {\pi \over 4} = \left| {{{m - 3} \over {1 + 3m}}} \right|$$

$$ \Rightarrow {{m - 3} \over {1 + 3m}} = \pm \,1$$

$$ \Rightarrow {{m - 3} \over {1 + 3m}} = + 1$$ and $${{m - 3} \over {1 + 3m}} = - 1$$

$$ \Rightarrow m - 3 = 1 + 3m$$ and $$m - 3 = - 1 - 3m$$

$$ \Rightarrow 2m = - 4$$ and $$4m = 2$$

$$m = - 2$$ and $$m = {1 \over 2}$$

We know, Equation of tangent to the parabola $${y^2} = 4m$$ is $$y = mx + {a \over m}$$ and point of contact is $$\left( {{a \over {{m^2}}},\,{{2a} \over m}} \right)$$

$$\therefore$$ Equation of tangent

$$y = - 2x - {a \over 2}$$

and $$y = {x \over 2} + 2a$$

$$\therefore$$ Point of contact A and B are

$$A\left( {{a \over {{{( - 2)}^2}}},{{2a} \over { - 2}}} \right) = A\left( {{a \over 4}, - a} \right)$$

$$B\left( {{a \over {{{\left( {{1 \over 2}} \right)}^2}}},{{2a} \over {\left( {{1 \over 2}} \right)}}} \right) = B\left( {4a,4a} \right)$$

As points A, B and S are colinear so area of triangle formed by those 3 points are zero.

Area of $$\Delta$$ABS = $${1 \over 2}\left| {\matrix{ {{a \over 4}} & { - a} & 1 \cr {4a} & {4a} & 1 \cr a & 0 & 1 \cr } } \right|$$

$$ = {a \over 4}(4a - 0) + a(4a - a) + 1(0 - 4{a^2})$$

$$ = {a^2} + 3{a^2} - 4{a^2} = 0$$

$$\therefore$$ 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.

Parabola

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