The equation x2 - Dy2 = 1 has a direct Diophantine connection to an ellipse for a special case.
The ellipse is: (D-1)j2 + (j +1)2 = (x+y)2 where j = ((x+Dy)-1)/D, when x = 1 mod D.
Or: (D-1)j2 + (j -1)2 = (x+y)2 where j = ((x+Dy)+1)/D, when x = -1 mod D.
So the special case is x = 1 mod D or -1 mod D, which must be true if D is a prime or twice a prime, but the connection will exist whenever x meets the conditions.
Using my method to reduce binary quadratic Diophantine equations on x2 - Dy2 = 1, gives:
(x+Dy)2 - D(x+y)2 = -(D-1)
(If you multiply out and simplify you simply return to x2 - Dy2 = 1.)
Shifting things around a bit, I have:
(x+Dy)2 + (D-1) = D(x+y)2
And I can group to divide off D:
[(x+Dy)2 - 1]/D + 1 = (x+y)2
Now we'll shift to the special case of x = 1 mod D or x = - 1 mod D.
Introducing integer j, I have x+Dy = jD +/- 1, and only one of those cases will be true. That usage is to keep from writing x+Dy = jD + 1 or x+Dy = jD - 1, over and over again.
And making that substitution gives:
[(jD +/- 1)2 - 1]/D + 1 = (x+y)2
And squaring and simplifying a bit gives:
Dj2 +/- 2j + 1 = (x+y)2
So now I just subtract and add j2 so that I can complete the square to get:
(D-1)j2 + (j +/- 1)2 = (x+y)2
And you have your ellipse.
Here is an example of the result with the famous case of D=61:
17663190492 - 61(226153980)2 = 1
x = 1766319049, y = 226153980, and x = 1766319049 = -1 mod 61, so the second equations apply.
(D-1)j2 + (j - 1)2 = (x+y)2, j = (x+Dy+1)/D
Which gives j = 255110030.
60*2551100302 + 2551100292 = 19924730292
If you'd like to continue further, additional research shows you can now generate yet another hyperbola from the ellipse.