That is

3C squared = — 4CD squared/3.
[Hence] C = 4D squared/9.

And equation III. becomes: 

4D squared        4D squared          16D^{4}
----- y squared - ----- x squared = - ---------
9          3             27

[TEX:  \frac{4D^2}{9} y^2 — \frac{4D^2}{3} x^2 = -\frac{16D^4}{27}]
____
/ 4D squared      2D
The semi-transverse axis = \/ ----- =  ----
9        3

[TEX:  \text{The semi-transverse axis} = \sqrt{\frac{4D^2}{9}}
= \frac{2D}{3}]
                               ____
                              / 4D squared 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

____________
4D squared     4D squared     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.

* * * * *

TEST CARD HINTS.

By Dr. F. Ogden stout.

I know it is the custom with a great many if not the majority of opticians to fit a customer without knowing whether he has presbyopia, hypermetropia, or any of the other errors of refraction.  Their method is first to try a convex, and if this does not improve, a concave, etc., until the proper one is found.  This, of course, amounts to the same thing if the right glass is found.  But in practice it will be found both time saving and more satisfactory to first decide with what error you have to deal.  It is very simple, and, where you have no other means of diagnosing (such as the ophthalmoscope), it does away with the necessity of trying so many lenses before the proper one is found.  You should have a distance test card placed at a distance of twenty feet from the person you are examining, and in a good light.