Find the equation of a parabola defined by two points and two tangents at those point A parabola passes through the point $(0 ,0)$ at an angle of $32$ degrees.
It also passes through the point $(0.312,0.117)$ at an angle of $7$ degrees.
How to find its equation?
 A: You will need to specify angle relative to what.
Here is how to solve it: let $y = f(x)$ be the functions whose graph satisfies the above properties.  Suppose that the parabola passes through the point $(a,f(a))$ with an angle $\alpha$ relative to the $x$-axis. The derivative of the function $y = f(x)$ whose graph is the parabola satisfies $f'(a) = \tan{\alpha}$.   Thus you are given
$$f'(0) = \tan(32)$$ and $$f'(0.312) = \tan(7).$$
Since $f$ is quadratic, $f'$ is linear. Thus $f'(x) = mx + b$ for some $m$ and $b$. You can use the equations above to work out $m$ and $b$.  Finally, since $f(0) = 0$ you will have $f(x) = \dfrac 12 mx^2 + bx$.
A: It's overdetermined, so you can throw one thing out, I suggest the angle of $7^\circ$ because it's ugly. You'll set up these equations:
$$f(0) = 0\\
f(0.312) = 0.117\\
f'(0) = \tan(32^\circ)$$
Where $f(x) = ax^2 + bx + c$ for $a,b$ and $c$ to be determined.
A: HINT: This is a rather simple Hermite interpolation problem, as you are effectively interpolating between two points with given values and derivatives. A textbook on numerical analysis or approximation theory would typically have such examples.
