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Say I have a parabola $y=ax-bx^2$ where $a,b>0$ and $y=cx-dx^2$ where $c,d>0$.

I would like to find some sort of relationship relating $c$ and $d$ with $a$ and $b$ such that the two parabolas intersect at right angles.

Now I can see that the usual route would work, with a bit of brute force. What I mean by the 'usual' route is by solving them simultaneously, acquiring the coordinates of their intersection point, substituting this into their gradient functions, letting the product be $-1$ and working from there.

However, as you can see, this is rather laborious. Is there a better way of deriving such a result?

I am thinking something to do with constructing a quadratic whose roots are the gradients of the tangents, then letting the product of roots be $-1$. However, I cannot think of how I could construct such a quadratic.

Trogdor
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  • When you say 'intersection point', do you include their intersection at the origin? – Semiclassical Sep 12 '15 at 00:42
  • I had intended on focusing on the non origin intersection point =) – Trogdor Sep 12 '15 at 00:52
  • The "brute force" route you sketch is not that bad. It ends up a not-too-ugly polynomial relation between the coefficients. – hmakholm left over Monica Sep 12 '15 at 01:07
  • can you go backwards? Say you have two Quadratic Bezier curves A and B. Each has three points, two of which are on the parabola and one of which is a 'control point'. Now, draw a line between the control point and each on-curve point, you have two tangents. One of A tangent lines must cross one of B tangent lines at a right angle. Work backwards from this to find the equations of the parabolas? Just a theory. – don bright Sep 12 '15 at 01:14
  • @donbright: That way it would seem to be nontrivial to ensure that both parabolas have vertical axes. – hmakholm left over Monica Sep 12 '15 at 01:18
  • you were right it was not trivial, transforming implicit coeffecients to bezier coordinates did not result in a simple situation. using 'brute force' in a computer algebra system (sympy.org) the intersection is at $x=(a-c)/(b-d)$, and the relation to guarantee perpendicularity appars to be $a - \frac{2 b \left(a - c\right)}{b - d} = - \frac{1}{c - \frac{2 d \left(a - c\right)}{b - d}}$. – don bright Sep 13 '15 at 22:02

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