Substituting this
value of E in equation II., and omitting accents--
We have
y squared - 3x squared + 2Dx - D squared/3 + D squared - 2Dx + 2D squared/3 = o
[Hence] y squared - 3x squared = - 4D squared/3
[Illustration: Fig I]
[Illustration: Fig II]
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 Cy squared - 3Cx squared = - 4CD squared/3. [III.]
When in this form the product of the coefficients of the x squared and y squared
terms should be equal to the remaining term.
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.
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