Find a parabola touches the line $y=x$ and $y=-x$ at $A(3,3)$ and $B(1,-1)$ 
If a parabola touches the line $y=x$ and $y=-x$ at $A(3,3)$ and $B(1,-1)$, then find the focus, axis of the parabola and its directrix.

What I thought: Since the 2 tangents are perpendicular,the origin must lie on the directrix and the line joining A and B is a focal chord. Don't know how to proceed from here..
 A: Let’s continue with the geometric construction that you started.  

Construct the parallelogram $AOBC$. The diagonal $OC$ is parallel to the parabola’s axis. We know that perpendicular tangents meet on the directrix, which is perpendicular to the parabola’s axis, so we now know its directrix $d$. Construct circles centered on $A$ and $B$ and tangent to $d$. Their intersection is the parabola’s focus. You can simply divide $\overline{AB}$ proportionally to find this point: $F = {r_BA+r_AB\over r_B+r_A}$. You can then use the point-point and point-line distance formulas to obtain a Cartesian equation for the parabola.
The construction in the preceding paragraph took advantage of the perpendicularity of the two tangents, but it’s quite easy to obtain a Cartesian equation of a parabola from any pair of points $P_0$ and $P_2$ on it and the tangents at those points: Let $P_1$ be the intersection of the tangents. A Bézier parameterization of the parabola is then $(1-t)^2P_0+2t(1-t)P_1+t^2P_2$. Eliminating $t$ gets you a Cartesian equation for it. In this case, the tangents intersect at the origin, so we have the parameterization $$x = 3(1-t)^2+t^2 = 4t^2-6t+3 \\ y=3(1-t)^2-t^2 = 2t^2-6t+3.$$ Eliminating $t$ produces the equation $$x^2-4xy+4y^2-12x+6y+9=0.$$
A: Parabola Property:The line joining the mid point of the focal chord and the point of intersection of the tangents drawn at the end points of that focal chord is parallel to the axis.(derivative of this property https://en.wikipedia.org/wiki/Parabola#Axis-direction).
Therefore the slope of the axis=[3-(-1)]/[3-1)]=2
Since we already have one point lying on the directrix (0,0) and the slope of the axis,we can write the equation of the directrix.
Therefore 
Equation of Directrix:- y=-x/2
Now we also know that Image of the focus in any tangent of the parabola lies on the directrix.Hence if we assume coordinates of the focus as say (h,k) then we can find the coordinates of the focus.Hence we can easily find the equation of the parabola.
Let the tangent be y-x=0 in which we find the image whose coordinates are (x,y)
(x-h)/-1=(y-k)/1=-2(k-h)/(1^2+1^2)
Therefore x=k and y=h
Hence h=-k/2
The focus also lies on the AB,hence focus lies on y-3=2(x-3)
Therefore k-3=2(h-3).Hence we can write the equation of axis as well.
A: The most general form of a parabola is $$Ax^2+Bxy + Cy^2 + Dx +Ey +F=0$$
where $$B^2 = 4AC \tag{1}$$.
Letting $(x,y)=(3,3)$, we get $$9A + 9B + 9C + 3D + 3E + F = 0 \tag{2}$$
Letting $(x,y)=(1, -1)$, we get $$A - B + C + D - E + F = 0 \tag{3}$$
\begin{align}
   Ax^2+Bxy + Cy^2 + Dx +Ey +F=0
   &\implies 2Ax + By + Bxy' + 2Cyy' + D + Ey' = 0 \\
   &\implies (Bx + 2Cy + E)y' + (2Ax + By + D) = 0
\end{align}
Letting $(x,y,y') = (3,3,1)$, we get $(3B + 6C + E) + (6A + 3B + D) = 0$,
which implies $$6A + 6B + 6C + D + E = 0 \tag{4}$$
Letting $(x,y,y') = (1,-1,-1)$, we get $-(B - 2C + E) + (2A - B + D) = 0$,
which implies $$2A - 2B + 2C + D - E = 0\tag{5}$$
Solving equations $(1)$ through $(5)$ and letting $A=1$, we get
$$(A,B,C,D,E,F) = (1, -4, 4, -12, 6, 9)$$
So, the equation of the parabola becomes
$$x^2 - 4xy + 4y^2 - 12x + 6y + 9 = 0$$
Added because of something that I found out later.
If $A=0$, then $B^2=4AC$ implies $B=0$ and the equation becomes $Cy^2 + Dx +Ey +F=0$.
If $A \ne 0$, then $A < 0$ implies 
$$(-A)x^2 +(-B)xy + (-C)y^2 + (-D)x + (-E)y + (-F) = 0$$
and $B^2=4AC \iff (-B)^2 = 4(-A)(-C)$
So we may as well assume that $A = a^2 > 0$. Then
\begin{align}
   Ax^2 + Bxy + Cy^2
   &= \dfrac{1}{4A}(4Ax^2 + 4ABxy + B^2y^2) \\
   &= \dfrac{1}{4A}(2Ax+By)^2 \\
   &= (ax+by)^2 \\
\end{align}
where $b = \dfrac{B}{2a}$.
So we can write the most general form of a parabola as $$(ax + by)^2 + Dx +Ey +F=0$$
We can now argue much as I did above and get the same answer without being annoyed by the nonlinear equation $B^2 = 4AC$.
