Let o d = D
[Hence] h d = D-x
Let c h = y
[Hence] (2x)² = y² + (D-x)²
or 4x² = y² + D²-2Dx + x²
[Hence] y²-3x² + D²-2Dx = o [I.]
This is the equation of an hyperbola whose center is on the axis of
abscisses. In order to determine the position of the center, eliminate
the x term, and find the distance from the origin o to a new origin
o'.
Let E = distance from o to o'
[Hence] x = x' + E
Substituting this value of x in equation I.
y²-3(x' + E)² + D²-2D(x' + E) = o
or y²-3x²-6Ex'-3E² + D²-2Dx'-2DE = o [II.]
In this equation the x' terms should disappear.
[Hence] -6Ex' - 2Dx' = o
[Hence] -E = - D/3
That is, the distance from the origin o to the new origin or the
center of the hyperbola o' is equal to one-third of the distance
from o to d; and the minus sign indicates that the measurement
should be laid off to the left of the origin o. Substituting this
value of E in equation II., and omitting accents--