3C² = - 4CD²/3.
[Hence] C = 4D²/9.
And equation III. becomes:
4D² 4D² 16D^{4}
----- y² - ----- x² = - ---------
9 3 27
[TEX: \frac{4D^2}{9} y^2 - \frac{4D^2}{3} x^2 = -\frac{16D^4}{27}]
____
/ 4D² 2D
The semi-transverse axis = \/ ----- = ----
9 3
[TEX: \text{The semi-transverse axis} = \sqrt{\frac{4D^2}{9}}
= \frac{2D}{3}]
____
/ 4D² 2D
The semi-conjugate axis = \/ ----- = -----
3 ___
\/ 3
[TEX: \text{The semi-conjugate axis} = \sqrt{\frac{4D^2}{3}}
= \frac{2D}{\sqrt{3}}]
Since the distance from the center of the curve to either focus is
equal to the square root of the sum of the squares of the semi-axes,
the distance from o' to either focus
____________
/4D² 4D² 4D
= /\ /----- + ----- = ----
\/ 9 3 3
[TEX: \sqrt{\frac{4D^2}{9} + \frac{4D^2}{3}} = \frac{4D}{3}]
We can therefore make the following construction (Fig. II.) Draw a d
the chord of the arc a c d. Trisect a d at o' and k. Produce
d a to l, making a l = a o' = o' k = k d. With a k as a
transverse axis, and l and d as foci, construct the branch of the
hyperbola k c c' c", which will intersect all arcs having the common
chord a d at c, c', c", etc., making the arcs c d, c' d, c"
d, etc., respectively, equal to one-third of the arcs a c d, a c' d,
a c" d, etc.
* * * * *