Finding equation of the parabola from tangent information Two lines drawn through $T$ at $(-1,-2)$ are tangent to a parabola at $P$ $(2,3)$ and $Q$ $(3,-1)$, respectively.  Find the equation of the parabola.
I tried using similarity of triangles to find the focus, but to no avail. 
 A: Method 1: First, construct the parallelogram $PTQR$; the diagonal $TR$ is parallel to the parabola’s axis. It’s convenient to use the midpoint of $PQ$ instead of constructing $R$ explicitly—diagonals of a parallellogram bisect each other. Next, use the reflective property to find the focus: reflect the line through $P$ parallel to $TR$ in the tangent line $TP$ and likewise reflect the line through $Q$ in the tangent line $TQ$. The focus is the intersection of the two reflections. You can now find the equation of the axis $ax+by+c=0$. The equation of the parabola will have the form $$(ax+by+c)^2 = p(by-ax+d),$$ with $p$ and $d$ yet to be determined. A straightforward way to find these unknown parameters is simply to plug in the coordinates of $P$ and $Q$ and solve the resulting system of equations. Alternatively, you can drop a perpendicular from the focus to either of the tangent lines. The foot of the perpendicular lies on the tangent to the vertex, which will let you determine its equation—and hence $d$—and then you can find $p$ by substitution.  
Method 2: Construct the Bézier parameterization of this parabola $(1-t)^2 P+2t(1-t)T+t^2Q$ and eliminate $t$. A mechanical way to do the latter is to compute the resultant of the two polynomials and factor $t$ out of it. The resulting equation will have to form of a general conic equation, which I find more opaque than the one developed in method 1.  

Method 1.1 (added): Once you’ve found a direction vector $(\beta,-\alpha)$ for the axis, you can construct the skeleton of an equation for the parabola: $$f(x,y) = (\alpha x+\beta y)^2+Dx+Ey+F=0.$$ The two known points $P$ and $Q$ give you two equations for the unknown coefficients. The normal and tangent at any point are orthogonal, so you can write $\nabla f(P)\cdot(P-T) = 0$ and $\nabla f(Q)\cdot(Q-T)=0$ for two more.
Method 1.2 (added): A classical definition of a parabola is the set of points equidistant from a fixed point (focus) and fixed line (directrix). Knowing the coordinates $(x_f,y_f)$ of the focus and that the equation of the directrix will have the form $by-ax+d=0$, you can use standard distance formulas to write the equation $$(x-x_f)^2+(y-y_f)^2 = {(bx-ay+d)^2\over a^2+b^2}$$ for the parabola. Now there’s only one unknown parameter $d$, which you can find by plugging the coordinates of either of the known points into this equation.
