Given a 2 points and the tangent slope at one of those points, how can I calculate a parabola? I am given $3$ bits of data:


*

*Point $A$ = $(-14 , 277)$

*Point $B$ = $(793 , 3)$

*The slope of the tangent at the point $A$ = $20°$


The only other data known about the parabola is that it aims down $-Y$ similarly to a ballistic trajectory.
What formulae are used to calculate said parabola?
 A: A parabola is of the form $$y=ax^2+bx+c.\tag1$$  The derivative is $$y'=2ax+b.\tag2$$  Substituting the two given points in $(1)$ and the derivative in $(2)$ gives three linear equations in the unknowns $a,b,c$.  Solve the system. 
A: $y = y_a + y'(x_a) (x- x_a) + \frac {y_b - y'(a)(x_b-x_a)}{(x_b-x_a)^2}(x - x_a)^2$
If $x = x_a$ then all of the $(x-x_a)$ factors equal $0$ and $y(x_a) = y_a$
$y'(x) = y'(x_a) + 2\frac {y_b - y'(a)(x_b-x_a)}{(x_b-x_a)^2}(x - x_a)$
and at $x = x_a$ then $y'(x_a) = y'(x_a)$
If $x = x_b$ then $y = y_b$
A: Say the parabola passes through $(x_1,y_1)$ and is tangent to a line $y=mx+b$ at $x_2,mx_2+b)$.  Given $x_2,y_2,m$ we may put  $b=y_2-mx_2$.  Then the difference between $y$ at any point $(x,y)$ on the parabola and the $mx+b$ on the line must be proportional to the square of the horizontal displacement from the tangency point:
$y=mx+b+k(x-x_2)^2$
Now we need only $k$, which is solved for by putting the other known point $(x_1,y_1)$:
$y_1=mx_1+b+k(x_1-x_2)^2$
$k=(y_1-mx_1-b)/(x_1-x_2)^2$
And then
$y=mx+b+((y_1-mx_1-b)/(x_1-x_2)^2)(x-x_2)^2$
