Fourth point on a parabola 
A upright parabola opening downwards passes through the following points:
  $$(0,0), (p,q), (q,p), (p+q, k)$$
  Find $k$ in terms of $p,q$. 

Of course one can always plug points into the standard parabola equation to find the coefficients and use that to find $k$. However, given the symmetry of the points, could there be a more elegant way of finding the solution? 

Added: Also is there a corresponding geometrical interpretation?

This problem was inspired by another problem here on arithmetic progressions. 
 A: That parabola passes through the origin means $x=0$, so it looks like $y=ax^2+bx$. Plugging the other 2 points and subtracting you get
$$
-(p-q) = a(p^2-q^2) + b(p-q)
$$
and now (assuming $p \ne q$, divide by $p-q$ to get
$$
-1 = a(p+q) + b
$$
which easily solves for $p+q$ without constructing the equation directly.
A: The geometric iterpretation is this. Draw a line which passes thru the 2nd and 3rd points. The four points are symmetric with respect to this line. The line intersects the $y$-axis at point $(0,p+q)$ and interesects the $x$-axis at point $(p+q,0)$. Since the 1st point $(0,0)$ is $p+q$ below the intersection at $x=0$, then the 4th point is $p+q$ below the intersection point $(p+q,0)$ and must be $(p+q,-p-q)$.
The symmetry is that the vertical line passing thru $(\frac{p+q}2,0)$ is the central axis of the parabola with respect to lines of slope $-1$. Such lines which intersect the parabola at two points also intersect the vertical line at a third point which is midway between the two parabola points.
