<s>[INST] 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..
 [/INST] Conditions (A),(B),(C),(D) are not needed to determine the tilted parabola ( $xy$  term non-zero). Because out of five constants needed to determine a conic, one can be reduced as zero determinant for parabola.You have given two points and two slopes which are quite sufficient.
$$ (x, y, y^{ \prime} ) = (3,3,1), (1,-1,-1) $$
Parabola involving $xy,x^2, y^2,x,y $ terms can be solved for y in a quadratic as:
$$ y^2 - 2y ( ax+b) + (ax+b)^2 - (cx+d) =0 $$
$$ y = (ax +b)  \pm \sqrt{ c x +d } $$
$\pm$ symbol separates regions on either side of vertical tangent. Plug in given point coordinates:
$$ 3 a + b+ \sqrt{   3c+d} =3 ,  \quad   a + b+ \sqrt{  c+d} = -1 ,\quad   a + c \ /( 2 \sqrt{   c+d} ) =  1  ,\quad   a - c \ /( 2 \sqrt{   c+d} ) = -1  $$
Using a CAS to reduce tedium we get $ (a,b,c,d) = ( 1/2, -3/4, 9/4,-27/16) $
The parabola is plotted to verify everything .It checks out (C), (D) but (A),(B) are inconsistent with first inputs and are clearly incorrect.
EDIT1
Some calculations needed (in progress)at focus and tangent orthogonal intersection on directrix
For parabola $ 4 a y = x^2 $ 
Points of tangency 
$$ (2at, a t^2)\quad (2a/t, a/t^2)  $$
Point of intersection of polar chord tangents on directrix which happens at right angles at a point D
$$ a [( t-1/t), -1] $$
Length of tangents $T_1,T_2$ given by
$$ (T_1/a)^2 =  t^4+ 3 t^2 + 1/t^2 +3 = 2, \quad  (T_2/a)^2 =  1/t^4+ 3/ t^2 + t^2 +3 = 18 $$
$$\rightarrow t= \frac13 $$
$OF$ is perpendicular on hypotenuse AB
$$ \frac{1}{OF^2} =\frac{1}{2} + \frac{1}{ 9 \cdot 2}  $$
$$ OF = \frac{3}{\sqrt 5} ... $$
$$ \cos \beta= \frac{OF}{OB} = \frac{ 3}{\sqrt10},\quad \sin \beta=  = \frac{ 1}{\sqrt10}\quad $$
to be continued 

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