Let g c = x
Therefore, from the definition c d = 2x
Let o d = D
[Hence] h d = D-x
Let c h = y
[Hence] (2x) squared = y squared + (D-x) squared
or 4x squared = y squared + D squared-2Dx + x squared
[Hence] y squared-3x squared + D squared-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.
y squared-3(x' + E) squared + D squared-2D(x' + E) = o
or y squared-3x squared-6Ex'-3E squared + D squared-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.
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