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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.

Hypergeometricx
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2 Answers2

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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.

gt6989b
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  • Nice solution - thanks (+1). There is a sign error - perhaps you could correct that. Could you also see if there might be a geometrical interpretation for the problem? – Hypergeometricx May 22 '17 at 16:23
  • @hypergeometric thanks for the note, sign error corrected. – gt6989b May 22 '17 at 16:57
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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.

Somos
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