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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.

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: $2alpha P_{Lx}+beta=a$.

You have then a system of three linear equations in three unknowns $alpha$, $beta$ and $gamma$.