Find the equation of a parabola and a point on the parabola given a two tangent lines. I've been trying to find a solution to this problem but I'm not too sure how to go about solving it.

I need to find the unknown values of $A, B$ and $C$ in the parabola
  equation $y(x) = Ax^2 + Bx + C$ given that the parabola passes through
  point (0,0) and is tangent to the line $y1(x) = 0.1x$ which also
  passes through the point (0,0).
I also need to find an unknown point (x-coordinate, y-coordinate) on
  the parabola which is tangent to $y2(x) = -0.08x + 10$ given that
  $y2(x) = -0.08x + 10$ passes through (200,-6).

I would greatly appreciate any help with solving this problem. So far I've tried to use the $y1(x)$ line to determine the vale of $B$ and $C$ in the parabola:
As the parabola passes through (0,0) I tried,
$$y(0) = 0$$ so 
$$A(0)^2 + B(0) + C = 0\\
C = 0$$
As the line $y1(x) = 0.1x$ is tangent to the parabola at (0,0) I tried,
$$ y'(0) = 0.1$$ so
$$ 2A(0) + B = 0.1\\               
B = 0.1$$
However when trying to find a point on the parabola where the parabola is tangent to $y2(x)$ it seems to me that the value for B would be different, so this has me really confused on how to solve this problem.
Once again, any help with this would be very very appreciated :)
 A: Hint:
Let the tangent meet the parabola at $(x_0,y_0)=(x_0,Ax_0^2+0.1x_0)$. The slope of the tangent is $$-0.08 = 2Ax_0+0.1, \tag 1$$
and the equation of the line implies $$Ax_0^2+0.1x_0 = -0.08x_0+10. \tag 2$$
Solve them simultaneously to compute $x_0$ and $A$.
A: Your condition means that $y=0.1x$ is the linear approximation of your quadratic polynomial at $x=0$. Thus the "error" made by replacing the one by the other is quadratic.
Otherwise said, you are looking for an equation of the form:
$$y=0.1x + \underbrace{A x^2}_{\text{error term}}$$
with a general parameter  $A$, with $B=0.1$ and $C=0$.
It is difficult for me to understand the second part of the problem.  Could you give more precision ?
A: Since it passes through $(0,0)$ then $C=0$. 
Since it has a tangent line on x=0, we know that its derivate must match the slope of the line it is tangent to. 
So we have $y(x) = ax^2 + bx$ 
$y'(x) = 2ax + b$ 
$y'(0) = b = 0.1$
So know we can write:
$y(x) = ax^2+0.1x$ 
We are still missing the value of $a$ we can use the next chunk of data to see if it resolves or not.
According the next piece of information the parabola needs to intercept somewhere $(x_0,y_0)$ the tangent $y_2(x)$ this means that:
$y_0=ax_0^2+0.1x_0$
Because the line needs to touch we know:
$y_2(x_0)=y_0 = ax_0^2+0.1x_0 =  -0.08x_0 +10$
This allows us to calculate the missing a based on the value of $x_0$. I however suspect the problem is not well written and probably they are giving you the tangent point as well when they mention $(200,6)$. 
In this case $x_0= 200$ and $y_0=6$ 
$a200^2+0.1\cdot200 = 6$ 
Which makes: 
$a=\frac{-7}{100^2}$
