Note:
@Gavran 's method above needs a second power form to take the discriminant.
@pasmith 's method works in the general case.
The two methods are otherwise very similar.
@Gavran sets both slopes the same and sets these to the slope of the normal line, just like
@pasmith .
@Gavran then translates one curve in the direction of the other, and has them make contact at one point.
@pasmith is a slightly easier route by not needing this extra step.
@pasmith (post 11) basically has ## y_1'(x_1)=y_2'(x_2)=-(\frac{x_2-x_1}{y_2-y_1}) ##. where the slopes are set the same and set equal to the slope of the normal line between them connecting ## (x_1,y_1) ## and ## (x_2,y_2) ##. This seems to be the simplest and a most general way to work problems of this kind.
@pasmith has this in his post for the most part, but doesn't spell it out in this simple form. He did a couple algebraic steps to get a result, in hindsight, we know that had to be the case.
@anuttarasammyak previously mentioned this as well, (post 26), but I thought I would write it out in its simplest form.
Looking over the posts, I actually spotted these two results in post 8, of the same slopes, and being perpendicular to the normal line between the points, but it wasn't until later, e.g.
@anuttarasammyak of post 26 that we started to see that these two results would indeed be a simple solution to the general case. :)