Q:

Transform parabola to be tangent to line in point and through other point

Sorry for stupid question, but I give up. I've spent whole weekend to solve that and no results. Please help. I think I know the solution, but it doesn't work form me. So now I am not sure. I am not sure if I make some stupid mistake when simplifying equations or just my method is wrong? But solving that is very time consuming, so before next try I decided to first ask you if my method is OK, or should I change something?
I have line: $f_1(x)=ax$
I have point on that line: $\ \ \ P _ { L x } $ ;$\ P _ { L y }$ 
I also have second point that is not on the line (somewhere on the plane): $\ \ \ P _ { x }$ ; $\ P _ { y }$ 
I also already have parabola that I need to transform to make it tangent to line $f_1(x)$, and through the points $P_L\ $ and $P$
My parabola equation is: $f _ { 2 } ( x ) = b x ^ { 2 } + b x$
So it seems to be obvious I need to find some unknown variables $A$, $B$ and $C$. Like that:
$f _ { 2 } ( x ) = A b ( x - B ) ^ { 2 } + A b ( x - B ) + C$
So my next step is:
$f _ { 1 } \left( x _ { 1 } \right) = f _ { 2 } \left( x _ { 1 } \right)$
which is: $A b \left( x _ { 1 } - B \right) ^ { 2 } + A b \left( x _ { 1 } - B \right) + C = a x _ { 1 }$
Then: $f _ { 1 } ^ { \prime } \left( x _ { 1 } \right) = f _ { 2 } ^ { \prime } \left( x _ { 1 } \right)$
which is: $2 A b \left( x _ { 1 } - B \right) + A b = a$
So now I have four unknown: $x_1$, $A$, $B$, $C$, and two equations, so I need two more.
So I use point $P_L$:
$A b \left( P _ { L x } - B \right) ^ { 2 } + A c \left( P _ { L y } - B \right) + C = P _ { L y }$
And point $P$:
$A b \left( P _ { x } - B \right) ^ { 2 } + A c \left( P _ { x } - B \right) + C = P _ { y }$
So now I have four equations:
$A b \left( x _ { 1 } - B \right) ^ { 2 } + A b \left( x _ { 1 } - B \right) + C = a x _ { 1 }$
$2 A b \left( x _ { 1 } - B \right) + A b = a$
$A b \left( P _ { L x } - B \right) ^ { 2 } + A c \left( P _ { L y } - B \right) + C = P _ { L y }$
$A b \left( P _ { x } - B \right) ^ { 2 } + A c \left( P _ { x } - B \right) + C = P _ { y }$
And four unknown: $x_1$, $A$, $B$, $C$
Is that OK???
It doesn't work for me. I can't solve that. I end up with very complicated equations with a lot of square roots with many strange expressions and can't find the way to simplify them. I tried wolframalpha and www.mathway.com but they also say it's too complicated for them. So maybe I do something wrong?
For any help great thanks in advance.

A:

I suppose you want to find the equation of a parabola, having its axis parallel to the $y$-axis, passing through a given point $P$ and touching a given line (of equation $y=ax$) at a given point $P_L$.
If so, the solution is quite straightforward. A parabola of that kind has a an equation of the form $y=\alpha x^2+\beta x+\gamma$ and passage through $P$ gives a first equation: $P_y=\alpha P_x^2+\beta P_x+\gamma$. Passage through point $P_L=(P_{Lx},aP_{Lx})$ gives a second equation: $aP_{Lx}=\alpha P_{Lx}^2+\beta P_{Lx}+\gamma$ and tangency at $P_L$ gives a third equation: $2\alpha P_{Lx}+\beta=a$.
You have then a system of three linear equations in three unknowns $\alpha$, $\beta$ and $\gamma$.

