Finding characteristic features of a parabola known by two points and their tangents This is from a math competition so it must not be something really long 
If a parabola touches the lines $y=x$ and $y=-x$ at $A(3,3) $ and $b(1,-1)$ respectively, then 
(A) equation of  axis of parabola is $2x+y=0$ 
(B)slope of tangent at vertex is $1/2$
(C) Focus is $(6/5,-3/5)$
(D) Directrix passes through $(1,-2)$ 
I thought the axis would be the angle bisector of the tangents passing through the focus but it turns out that is not the case in a parabola so how can I find anything..

(see figure below) 
Parametric equations for the parabola are readily obtained, assuming a certain knowledge of quadratic Bezier curves : see for example the paragraph "Second order curve is a parabolic segment" in (https://en.wikipedia.org/wiki/B%C3%A9zier_curve)).
The tangents' intersection is clearly $O(0,0)$. Thus the parabola is nothing but the Bezier curve with endpoints $A$ and $B$ and "directing" point in $O$: $M=A(1-t)^2 + 2t(1-t)O + B t^2$, otherwise said:
$$\tag{1}\pmatrix{x\\y}=(1-t)^2\pmatrix{3\\3}+2t(1-t)\pmatrix{0\\0}+t^2\pmatrix{1\\-1}$$
$$\tag{2}\Leftrightarrow \ \ \ \ \cases{x=3(1-t)^2+t^2\\y=3(1-t)^2-t^2}$$
which constitutes a parametric description of the curve. 
Eliminating $t$ between the two equations (2), one gets an implicit equation that can be written under the form :
$$\tag{3} \left(x-\frac{6}{5}\right)^2+\left(y+\frac{3}{5}\right)^2=\frac{(y+2x)^2}{5}$$
Taking square roots on both sides:
$$\tag{4} \sqrt{\left(x-\frac{6}{5}\right)^2+\left(y+\frac{3}{5}\right)^2}=\frac{|y+2x|}{\sqrt{5}}$$
(4) expresses the fact that the distance of $M(x,y)$ (the current point on the curve) to point $F(\frac{6}{5},-\frac{3}{5})$ is equal to the distance of $M$ to straight line $(D)$ with equation $y+2x=0$ (with slope $-2$). For distance formula see: (http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html). Thus $F$ is the focus of the parabola and $(D)$ its directrix. 
Therefore, the parabola's axis, passing through $F$ and orthogonal to $(D)$ (thus with slope $\frac{1}{2}$) has the following equation: $y+\frac{3}{5}=\frac{1}{2}(x-\frac{6}{5})$, i.e., $y=\frac{1}{2}x-\frac{6}{5}.$
Conclusion:


*

*propositions (A), (B) are false. For (B), it's not necessary to compute the tangent at the vertex because the slope of this tangent is the same as the slope of the directrix, which is $-2$, not $\frac{1}{2}$.

*propositions (C), (D) are exact.

