This is Right!

Let the points P & Q be $(at^{2}_{1},2at_{1})$ & $(at^{2}_{2},2at_{2})$ .

Here $a$ $=$ $\frac{1}{4}$ ,

Therefore the points are P $(\frac{t^{2}_{1}}{4},\frac{t_{1}}{2})$ & Q $(\frac{t^{2}_{2}}{4},\frac{t_{2}}{2})$

So the point of contact of the tangents is $(at_{1}t_{2},a(t_{1} + t_{2}))$

So, point of contact is $(\frac{t_{1}t_{2}}{4},\frac{t_{1}+t_{2}}{4})$

As the point of contact lies on the line $x$ $+$ $1$ $=$ $0$

So $t_{1}t_{2}$ $=$ $-4$

If $t_{1}$ $=$ $t$ then $t_{2}$ $=$ $\frac{-4}{t}$

So the points are P $(\frac{t^{2}}{4},\frac{t}{2})$ & Q $(\frac{4}{t^{2}},\frac{-2}{t})$

So the area of the triangle is given by,

$\frac{1}{2}$ ( $\begin{vmatrix} x_{1} & y_{1} \\ x_{2} & y_{2}\end{vmatrix}$ $+$ $\begin{vmatrix} x_{2} & y_{2} \\ x_{3} & y_{3}\end{vmatrix}$ $+$ $\begin{vmatrix} x_{3} & y_{3} \\ x_{1} & y_{1}\end{vmatrix}$ )

So by substituting the values we get,

Area $=$ $\frac{1}{t}$ $+$ $\frac{t}{4}$

So area is maximum when $t= 2$ .

So the area is equal to $1$ .

So $4M$ $=$ $4$