This is the equation of an hyperbola referred to its center o' as the origin of co-ordinates. To write it in the ordinary form, that is in terms of the transverse and conjugate axes, multiply each term by C, i.e., __ Let \/C = semi-transverse axis. [TEX: \sqrt{C} = \text{semi-transverse axis.}] Thus Cy² - 3Cx² = - 4CD²/3. [III.] When in this form the product of the coefficients of the x² and y² terms should be equal to the remaining term. That is 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. * * * * * TEST CARD HINTS. By Dr. F. OGDEN STOUT.