I believe there is a way to geometrically interpret integrating $R(x,\sqrt{Ax^2+Bx+C})$, as a means to motivate the Euler substitutions, in terms of

"...expressing the coordinates of a point upon a conic $y^2 = Ax^2+Bx+C$ by means of rational functions of a parameter. It can be seen geometrically that this is possible,. For, if a secant $y - b = t(x - a)$ be drawn through any point $(a,b)$ on the conic, the coordinates of the second point of intersection of the secant with the conic are given by equations of the first degree, and are therefore rational functions of $t$".

Goursat - Page 215

Can anyone parse this for me? I don't really see how this shows the second point of intersection is a rational function of the first degree.

This interpretation allows one to geometrically see that if the quadratic has imaginary roots, $A > 0$ and so this immediately implies the quadratic is a hyperbola. Therefore, geometrically, we see our 'Euler substitution' simply must be $y = \sqrt{A}x + t$, a straight line parallel to an asymptote of the hyperbola, and because it's an asymptote it cuts the hyperbola at the point

$$x = \frac{C-t^2}{2\sqrt{A}t-2B}, y = t+\sqrt{A}\frac{C-t^2}{2\sqrt{A}-2B}.$$

Similarly, if $A < 0$ the conic simply must be an ellipse, otherwise the quadratic is negative, and so geometrically we see we should substitute $y=t(x-a)$, the moving secant cutting the conic.

I believe this is an answer to this question, however I'd like somebody to parse all this out in a clearer form for us all to read, i.e. to make some of it more concrete and mathematical.

If one is masochistic enough, they could relate it to this interesting post also, but that's not necessary, thank you

  • $\begingroup$ In your quoted paragraph the letters $a,b$ seem to be doing two different jobs. $\endgroup$ May 19, 2014 at 21:37
  • $\begingroup$ Apologies, fixed. $\endgroup$
    – bolbteppa
    May 19, 2014 at 21:44
  • $\begingroup$ Your answer seems perfectly clear to me. So I dont quite understand what more are you looking for ? $\endgroup$ May 20, 2014 at 1:23


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