The value 3 is a consequence of a different method to draw the curve, namely not by converting it to parameter form but rather by recursively refining it.
That is, staring from four points $A,B,C,D$ you take successive mid points $E=\frac{A+B}2$, $F=\frac{B+C}2$, $G=\frac{C+D}2$, then midpoints of the midpoints $H=\frac{E+F}2$, $I=\frac{F+G}2$, finally $J=\frac{H+I}2$. Note that taking mid pints is a very simple operation with respect to the coordinates and may require much less precision than drawing the parametric curve.
Now the crucial point is the following: The curve determined by the four points $A,B,C,D$ is replaced by two smaller pieces of curve, the first determined by $A,E,H,J$, the other by $J,I,G,D$. Several nice properties are obeyed by this replacement so that the rough shape of the curve can be "predicted": The curve passes through end points ($A$ and $D$ of the original curve, additionally $H$ for the refined curves), the tangents there point towards $B$ resp. $C$.
Also, if the quadrangle $ABCD$ is convex, then the curve is contained inside it. Note that the refinement steps will sooner or later produce convex quadrangles and each refinement step from then on will give better and better containment estimates for the curve.
Ultimately, it is possible to calculate what the curve described by the above procedure sould look like in parametric form and it turns out that the factor 3 you observed comes into play. In effect, this is the same $3$ as in the binomial formula $(a+b)^3=a^3+3a^2b+3ab^2+b^3$. (Which means that for quadratic Bezier curves you will find a factor of $2$ and for Bezier curves of degree $4$, the numbers $4$ and $6$ will occur etc.
I think that I understand your confusion. Let me know if this works.
For a Bezier curve defined as you did, you have a parametric form:
$$x(t)=(1-t)^3\cdot 0+3t(1-t)^2\cdot .2+3t^2(1-t)\cdot.5+t^3\cdot 1$$
and
$$y(t)=(1-t)^3\cdot 0+3t(1-t)^2\cdot .5+3t^2(1-t)\cdot.9+t^3\cdot 1$$
When you plug in $t=1/2$, you get the point $(0.3875,0.65)$, which is not the point that you're looking for.
I think that you're interested in the case where $x=\frac{1}{2}$. Using Maple, I solved $x(t)=\frac{1}{2}$, and found that this happens when $t\approx0.60969549401666900$.
Plugging this $t$ value into $y(t)$ results in $\approx0.7576964125$, which, I think, is the value you're looking for.
Best Answer
If you know the value of the four points, then the parametric equation for curve P is found through the following: $$\left(\left(1-t\right)\left(\left(1-t\right)\left(\left(1-t\right)x_{1}+tx_{2}\right)+t\left(\left(1-t\right)x_{2}+tx_{3}\right)\right)+t\left(\left(1-t\right)\left(\left(1-t\right)x_{2}+tx_{3}\right)+t\left(\left(1-t\right)x_{3}+tx_{4}\right)\right),\left(1-t\right)\left(\left(1-t\right)\left(\left(1-t\right)y_{1}+ty_{2}\right)+t\left(\left(1-t\right)y_{2}+ty_{3}\right)\right)+t\left(\left(1-t\right)\left(\left(1-t\right)y_{2}+ty_{3}\right)+t\left(\left(1-t\right)y_{3}+ty_{4}\right)\right)\right)$$ where the domain of $t$ is $[0,1]$ This gives us $$x_{t}=x_{1}-3\left(x_{1}-x_{2}\right)t+3t^{2}\left(x_{1}-2x_{2}+x_{3}\right)-t^{3}\left(x_{1}-3x_{2}+3x_{3}-x_{4}\right)$$ $$y_{t}=y_{1}-3\left(y_{1}-y_{2}\right)t+3t^{2}\left(y_{1}-2y_{2}+y_{3}\right)-t^{3}\left(y_{1}-3y_{2}+3y_{3}-y_{4}\right)$$ You want the $y=mx+b$ formula for the line passing through the point $(x_a,y_a)$. Any $(x_a,y_a)$ value obtained through the above formula has a corresponding slope given by the set of parametric equations $x_t=x_t$ and $y'_t=\frac{\frac{d}{dt}{y_t}}{\frac{d}{dt}{x_t}}$ This gets us the following: $$x_{t}=x_{1}-3\left(x_{1}-x_{2}\right)t+3t^{2}\left(x_{1}-2x_{2}+x_{3}\right)-t^{3}\left(x_{1}-3x_{2}+3x_{3}-x_{4}\right)$$ $$y'_{t}=\frac{-\left(3\left(y_{1}-y_{2}\right)-6t\left(y_{1}-2y_{2}+y_{3}\right)+3t^{2}\left(y_{1}-3y_{2}+3y_{3}-y_{4}\right)\right)}{-\left(3\left(x_{1}-x_{2}\right)-6t\left(x_{1}-2x_{2}+x_{3}\right)+3t^{2}\left(x_{1}-3x_{2}+3x_{3}-x_{4}\right)\right)}$$ $$y'_{t}=\frac{\left(y_{1}-y_{2}\right)-2t\left(y_{1}-2y_{2}+y_{3}\right)+t^{2}\left(y_{1}-3y_{2}+3y_{3}-y_{4}\right)}{\left(x_{1}-x_{2}\right)-2t\left(x_{1}-2x_{2}+x_{3}\right)+t^{2}\left(x_{1}-3x_{2}+3x_{3}-x_{4}\right)}$$ This gives you the slope of the line. With a point and a slope, we can develop our linear equation. $$y-y_{t}=y'_{t}\left(x-x_{t}\right)$$ $$y=y'_{t}x-y'_{t}\left(x_{t}\right)+y_{t}$$ $$y=\frac{\frac{d}{dt}y_{t}}{\frac{d}{dt}x_{t}}x-\frac{\frac{d}{dt}y_{t}}{\frac{d}{dt}x_{t}}x_{t}+y_{t}$$ But if you want to brute force an equation, here is an expanded version of that. $$y=\frac{\left(y_{1}-y_{2}\right)-2t\left(y_{1}-2y_{2}+y_{3}\right)+t^{2}\left(y_{1}-3y_{2}+3y_{3}-y_{4}\right)}{\left(x_{1}-x_{2}\right)-2t\left(x_{1}-2x_{2}+x_{3}\right)+t^{2}\left(x_{1}-3x_{2}+3x_{3}-x_{4}\right)}x-\frac{\left(y_{1}-y_{2}\right)-2t\left(y_{1}-2y_{2}+y_{3}\right)+t^{2}\left(y_{1}-3y_{2}+3y_{3}-y_{4}\right)}{\left(x_{1}-x_{2}\right)-2t\left(x_{1}-2x_{2}+x_{3}\right)+t^{2}\left(x_{1}-3x_{2}+3x_{3}-x_{4}\right)}\left(x_{1}-3\left(x_{1}-x_{2}\right)t+3t^{2}\left(x_{1}-2x_{2}+x_{3}\right)-t^{3}\left(x_{1}-3x_{2}+3x_{3}-x_{4}\right)\right)+y_{1}-3\left(y_{1}-y_{2}\right)t+3t^{2}\left(y_{1}-2y_{2}+y_{3}\right)-t^{3}\left(y_{1}-3y_{2}+3y_{3}-y_{4}\right)$$ At any value t from 0 to 1, that will be your tangent line passing through the point $(x_{1}-3\left(x_{1}-x_{2}\right)t+3t^{2}\left(x_{1}-2x_{2}+x_{3}\right)-t^{3}\left(x_{1}-3x_{2}+3x_{3}-x_{4}\right),y_{1}-3\left(y_{1}-y_{2}\right)t+3t^{2}\left(y_{1}-2y_{2}+y_{3}\right)-t^{3}\left(y_{1}-3y_{2}+3y_{3}-y_{4}\right))$