Whence it is seen that DC is an ellipse, of which the axis DO is to the parameter as 9 to 5; and therefore the square on DO is to the square of the distance between the foci as 9 to 9 - 5, that is to say 4; and finally the line DO will be to this distance as 3 to 2.
[Illustration] Again, if one supposes the point B to be infinitely distant, in lieu of our first oval we shall find that CDE is a true Hyperbola; which will make those rays become parallel which come from the point A.And in consequence also those which are parallel within the transparent body will be collected outside at the point A.Now it must be remarked that CX and KS become straight lines perpendicular to BA, because they represent arcs of circles the centre of which is infinitely distant.
And the intersection of the perpendicular CX with the arc FC will give the point C, one of those through which the curve ought to pass.

And this operates so that all the parts of the wave of light DN, coming to meet the surface KDE, will advance thence along parallels to KS and will arrive at this straight line at the same time; of which the proof is again the same as that which served for the first oval.

Besides one finds by a calculation as easy as the preceding one, that CDE is here a hyperbola of which the axis DO is 4/5 of AD, and the parameter equal to AD.

Whence it is easily proved that DO is to the distance between the foci as 3 to 2.
[Illustration] These are the two cases in which Conic sections serve for refraction, and are the same which are explained, in his _Dioptrique_, by Des Cartes, who first found out the use of these lines in relation to refraction, as also that of the Ovals the first of which we have already set forth.

The second oval is that which serves for rays that tend to a given point; in which oval, if the apex of the surface which receives the rays is D, it will happen that the other apex will be situated between B and A, or beyond A, according as the ratio of AD to DB is given of greater or lesser value.