I would like to find the equation of a downward-facing parabola given a starting point (x,y) and angle L, and the height of the vertex (k).
I started with three equations:
$y = ax^2 + bx + c$ (equation of a parabola)
$y = a(x – h)^2 + k$ (equation for the vertex)
$\tan(L) = 2ax + b$ (to solve for the angle L)
In the picture, green variables are known, red variables are unknown, and I'd like to be able to solve for orange variables later once a, b, and c are known.
I tried to plug the three equations into each other a couple of different ways to solve for a, b, and c so I could solve for additional points on the parabola but the equations always turned out messy. So, I'm here to ask the internet for help!
I would like to be able to put the final solution into Excel so I can change the variables and update the final answer (which shouldn't be hard); I have a feeling it should also be possible to change the parabola to upward-facing by changing the sign of angle L.
Thanks in advance!!!
Best Answer
If $f(x) = ax^2 + bx + c$ is the equation of the parabola (with $a<0$), then the coordinates of the vertex are $M\left(-\frac{b}{2a}, -\frac{D}{4a}\right)$, where $D = b^2 - 4ac$. Thus, we have $$h = -\frac{b}{2a} \text{ and } k = -\frac{b^2-4ac}{4a}.$$
If $M(x_0,y_0)$ is the given point and $L$ is the angle as described, then we have that $$f'(x_0) = \tan L \iff 2ax_0 + b = \tan L $$
All this information should be sufficient to define $a,b,c.$
Thus, we have 3 equations (2 of them are linear) with 3 variables, namely $a,b,c$ and $x_0,y_0,L,k$ known quantities. $$\left.\begin{array}{rlr} ax_0^2 + bx_0 + c &= y_0 &\quad\quad \text{(linear)}\\ 2ax_0 + b &= \tan L&\quad \quad \text{(linear)}\\ 4ac-b^2 &= 4ak & \quad\quad\text{(non-linear)} \end{array} \right\} $$
This approach should simplify things in a way.