本帖最后由 hbghlyj 于 2021-12-23 22:58 编辑 以下内容是从古登堡项目的TeX文件转换来的.图片是SVG格式. Chapter Ⅰ Problem proposed by Cramer to Castillon
本章旨在将解析几何应用于历史上的一些著名问题。第一个是 M. Cramer 向 M. de Castillon 提出的，可以这样表述：给定三个点和一个圆，在圆中作一个内接三角形，该三角形的边分别通过给定的点。 (produce有"将(线段)延长"的意思) Let A, B, C be the given points. Draw a pair of
tangents from A, and let PQH be the line of contact.
Similarly pairs of tangents from B and C,
SRK, VTL being lines, of contact. Then if a
triangle KLH can be described about the circle,
and such that its angular points may be in the given
lines PQH, SRK, VTL respectively, then the
points of contact, X, Y, Z being joined will pass respectively
through A, B, C. For H being the pole
of ZX, tangents drawn where any line HQP intersects
the circle will intersect in ZX produced,
but those tangents intersect in A, and therefore
ZX passes through A. Similarly of the rest.
When any of the points A, B, C falls within the
circle as at $a$, join $oa$. Make $ap ⊥ oam$, and
draw tangent $pm$, then $\mathrm{A}m\mathrm{H} \perp om$ will hold the
place of PQH in the above.
We have therefore reduced the problem to the
following.
Let there be three given straight lines and a given
circle, it is required to find a triangle circumscribed
about the circle, which shall have its angular points
each in one of the three lines.
Let $a$ be the radius of the circle, and let the
equations to the required tangents be
\[
\left.
\begin{array}{l}
l_1 x + m_1 y = a \\
l_2 x + m_2 y = a \\
l_3 x + m_3 y = a
\end{array}
\right\} \tag{1}
\]
Also the equations to the three given lines
\[
\left.
\begin{aligned}
\mathrm{A}_1 x + \mathrm{B}_1 y &= p_1 \\
\mathrm{A}_2 x + \mathrm{B}_2 y &= p_2 \\
\mathrm{A}_3 x + \mathrm{B}_3 y &= p_3
\end{aligned}
\right\}
\tag{2}
\]
$p_1$, $p_2$, $p_3$ being perpendiculars upon them from the
centre of the circle, and $l_1$, $m_1$, $\mathrm{A}_1$, $\mathrm{B}_1$ etc., direction
cosines.
Suppose the intersection of the two first lines of
(1) to be in the third line of (2), we have by eliminating
$x$ and $y$ between
\begin{align*}
l_1 x + m_1 y &= a \\
l_2 x + m_2 y &= a \\
\text{and } \mathrm{A}_3 x + \mathrm{B}_3 y &= p_3
\end{align*}
the condition
\begin{align*}
\mathrm{A}_3 a(m_2 - m_1) + \mathrm{B}_3 a(l_1 - l_2) &= p_3(l_1 m_2 - l_2 m_1)
\end{align*}and similarly,\begin{align*}
\mathrm{A}_2 a(m_1 - m_3) + \mathrm{B}_2 a(l_3 - l_1) &= p_2(l_3 m_1 - l_1 m_3) \\
\mathrm{A}_1 a(m_3 - m_2) + \mathrm{B}_1 a(l_2 - l_3) &= p_1(l_2 m_3 - l_3 m_2)
\end{align*}
\begin{flalign*}
&\text{Now let }
&l_1 &= \cos \theta_1,\quad \therefore m_1 = \sin \theta_1 \text{ etc.} &&\\
&\text{Also }
&\mathrm{A}_3 &= \cos \alpha_3,\quad \therefore \mathrm{B}_3 = \sin \alpha_3 \text{ etc.} &&
\end{flalign*}
Then the first of the above conditions is
\[
a\{ \cos \alpha_3 (\sin \theta_2 - \sin \theta_1)
+ \sin \alpha_3 (\cos \theta_1 - \cos \theta_2) \}
= p_3 \sin (\theta_2 - \theta_1)
\]
This equation is easily reducible by ordinary trigonometry to
\[
\tan\frac{\alpha_3-\theta_1}{2} \tan\frac{\alpha_3-\theta_2}{2} +
\frac{p_3-a}{p_3+a} = 0
\]
Similarly
\begin{align*}
\tan \frac{\alpha_2-\theta_3}{2} \tan\frac{\alpha_2-\theta_1}{2} +
\frac{p_2-a}{p_2+a} &= 0
\\
\tan \frac{\alpha_1-\theta_2}{2} \tan\frac{\alpha_1-\theta_3}{2} +
\frac{p_1-a}{p_1+a} &= 0
\end{align*}
If now for brevity we put
$x = \tan\dfrac{\theta_1}{2}$,
$y = \tan\dfrac{\theta_2}{2}$,
$z=\tan\dfrac{\theta_3}{2}$, also
$k_3 =\dfrac{p_3+a\cos \alpha_3}{p_3+a\cos \alpha_3}$,
$h_3 = \dfrac{a\sin \alpha_3}{p_3-a \cos \alpha_3}$
etc. the above equations become
\begin{align*}
k_3xy -h_3(x+y)+1 & = 0 \\
k_2zx -h_2(z+x)+1 & = 0 \\
k_1yz -h_1(y+z)+1 & = 0
\end{align*}
from which we can immediately deduce a quadratic
for $x$.
On eliminating $z$ between the second and third
equations, we shall have another equation in $x$ and $y$
similar in form to the first.
We may, moreover, so assume the axis from which
$\alpha_1$, $\alpha_2$, $\alpha_3$ are measured, so that $h_3 = 0$ and the
equations are then,
\[
k_3 x y + 1 = 0
\]
\[
\text{and } \qquad (h_2 k_1 - h_1 k_2) x y + (h_1 h_2 - k_1) y -
(h_1 h_2 - k_2) x + h_2 - h_1 = 0.
\]
These are, considering $x$ and $y$ as co-ordinates,
the equations to two hyperbolas having parallel
asymptotes, and which we may assume to be rectangular.
To show that their intersections may be
easily determined geometrically, assume the equations
under the form
\begin{align*}
&x y = \mathrm{C}^2 \\
x y - \mathrm{C}^2 + &\mu \left( \frac{x}{\mathrm{A}} + \frac{y}{\mathrm{B}} - 1 \right) = 0
\end{align*}
Then by subtraction,
\[
\frac{x}{\mathrm{A}} + \frac{y}{\mathrm{B}} - 1 = 0
\]
is the common secant.
Let $\frac{x}{\mathrm{B}} + \frac{y}{\mathrm{A}} - 1=0$ be another secant.
Multiply these together and we have
\[
\frac{x^2 + y^2}{\mathrm{AB}} + \frac{\mathrm{A}^2 + \mathrm{B}^2}{\mathrm{A}^2\mathrm{B}^2} x y
+ \frac{\mathrm{A}+\mathrm{B}}{\mathrm{AB}} (x+y) + 1 = 0
\]
This equation represents the two secants. But at
the points of their intersection with the hyperbola
$xy= \mathrm{C}^2$, this last equation reduces to
\begin{align*}
x^2+y^2-&(\mathrm{A}+\mathrm{B})(x+y)+\mathrm{A}\mathrm{B}
+\mathrm{C}^2\frac{\mathrm{A}^2+\mathrm{B}^2}{\mathrm{A}\mathrm{B}}=0\\
\text{or}\quad &\left(x-\frac{\mathrm{A}+\mathrm{B}}{2}\right)^2
+\left(y-\frac{\mathrm{A}+\mathrm{B}}{2}\right)^2=(\mathrm{A}^2+\mathrm{B}^2)\left\{\frac{\mathrm{A}\mathrm{B}-
\mathrm{C}^2}{2\mathrm{A}\mathrm{B}}\right\}
\end{align*}
which represents a circle, co-ordinates of the centre
\begin{align*}
&x = y = \frac{\mathrm{A} + \mathrm{B}}{2} \\
\text{and radius } \quad
&(\mathrm{A}^2 + \mathrm{B}^2)^\frac{1}{2}
\left\{ \frac{\mathrm{AB}-\mathrm{C}^2}{2\mathrm{AB}} \right\}^\frac{1}{2}
\end{align*}
Hence it is evident that A and B, being once
geometrically assigned, the rest of the construction
is merely to draw this circle, which will intersect
$\dfrac{x}{\mathrm{A}} + \dfrac{y}{\mathrm{B}} - 1 = 0$ in the required points.
The analytical values of A, B, and C$^2$ are
\[
\begin{aligned}
\mathrm{A} &= - \frac{(h_1-h_2)k_3 + h_2 k_1 - h_1 k_2}{(h_1 h_2 - k_2 )k_3} \\
\mathrm{B} &= \frac{(h_1-h_2)k_3 + h_2 k_1 - h_1 k_2}{(h_1 h_2 - k_1 )k_3}\\
\mathrm{C}^2 &= -\frac{1}{k_3}
\end{aligned}\;
\]
These being rational functions of known geometrical
magnitudes, are of course assignable
geometrically, so that every difficulty is removed,
and the mere labour of the work remains.
本帖最后由 hbghlyj 于 2021-12-23 22:28 编辑 Tangencies of Apollonius
接下来,我用解析几何推导出阿波罗尼奥斯问题的可以适用于各种情况的一般构造模式.问题可以这样表述:给定三个圆(包括了退化的圆:点、直线),求作一个圆与给定的圆相切(对于退化情形:通过给定的点、与给定的直线相切).解决这个问题需要作出两条双曲线的交点,为此,需要应用以下两个命题:
(citerior directrix是指,所给的焦点对应的那条准线) The first proposition is, “If two conic sections have the same focus, lines
may be drawn through the point of intersection of
their citerior directrices,
and through two of the
points of intersection of the curves.”
Let $u$ and $v$ be linear functions of $x$ and $y$, so
that the equations $u = 0$, $v = 0$ may represent the
citerior directrices, then if $r = \sqrt{x^2 + y^2}$, and $m$
and $n$ be constants, we have for the equations of
the two curves
\begin{align*}
r &= m u \\
r &= n v
\end{align*}
and by eliminating $r$, $mu - nv = 0$; but this is
the equation to a straight line through the intersection
of $u = 0$, $v = 0$, since it is satisfied by these
simultaneous equations.
When the curves are both ellipses they can intersect
only in two points, and the above investigation
is fully sufficient. But when one or both the curves
are hyperbolic, we must recollect that only one
branch of each curve is represented by each of the
above equations. The other branches are,
\begin{align*}
r &= -mu \\
r &= -nv
\end{align*}
We have therefore, in this instance $mu + nv = 0$
as well as $mu - nv = 0$, for a line of intersection.
The second proposition is, having given the
focus, citerior directrix, and eccentricity of a conic
section, to find by geometrical construction the two
points in which the conic section intersects a given
straight line.
In either of the diagrams, the first of which is for
an ellipse, the second for a hyperbola, let MX be
the given straight line, F the focus, A
the vertex,
and DR the citerior directrix. Let
\[
\mathrm{FM + MX} = p, \quad \mathrm{MFD} = \alpha,
\]
$r$ the distance of any point in MX from F,
$\theta$ the
angle it makes with FD, and $\mathrm{FD} = a$.
Also let $n = \dfrac{\mathrm{FA}}{\mathrm{AD}}$
Then $\dfrac{r}{a-r \cos \theta}= n, \quad r \cos (\theta - \alpha) = p$
Eliminate $r$
\[
\frac{a}{p} \cos (\theta - \alpha) - \cos \theta = \frac{1}{n}
\]
or
\[
(a \cos \alpha - p) \cos \theta + a \sin \alpha \sin \theta = \frac{p}{n}
\]
Let
\[
\frac{a \cos \alpha - p}{a \sin \alpha} = \cot \varepsilon
\tag{1}
\]
Then
\[
\frac{a \sin \alpha}{\sin \varepsilon}
\cos (\theta - \varepsilon) = \frac{p}{n}
\]
or
\[
\cos (\theta - \varepsilon)
= \frac{p \sin \varepsilon}{n a \sin \alpha} = \frac{p}{nd}
\tag{2}
\]
where $d = \dfrac{a \sin \alpha}{\sin \varepsilon}$.
From the formulae (1) and (2) we derive the following
construction. Join DM, then DMH is
the angle $\varepsilon$, because DM projected on FH is $a \cos \alpha - p$. Also a perpendicular from D on FH
is $a \sin \alpha$,
$\therefore \dfrac{a \cos \alpha - p}{a \sin \alpha} = \cot \mathrm{DMH}$,
$\therefore \cot \mathrm{DMH} = \cot \varepsilon$,
$\therefore \mathrm{DMH} = \varepsilon$
Again, $ \dfrac{\sin \mathrm{DMF}}{\sin \mathrm{DFM}} = \dfrac{\mathrm{DF}}{\mathrm{DM}} $, or
$\dfrac{\sin \varepsilon}{\sin \alpha} = \dfrac{a}{\mathrm{DM}}$, whence
$\mathrm{DM} = d$. Find ML a third proportional to
AD, FA and DM, so that $\mathrm{ML} = nd$. With
centre M and radius ML describe a circle. Make
MH equal to FM, and draw KHL at right angles
to FH, and join MK, ML. Then by (2) LMH
or $\mathrm{KMH} = \theta - \varepsilon$.
Taking the value $\varepsilon - \theta$, we have therefore
\[
\mathrm{LMD}= \mathrm{DMH} - \mathrm{LMH} = \varepsilon -(\varepsilon - \theta) = \theta.
\]
And taking $\theta - \varepsilon$, we have
\[
\mathrm{KMD} = \mathrm{KMH} + \mathrm{HMD} = \varepsilon + \theta - \varepsilon = \theta.
\]
Hence, make $\mathrm{QFX} = \mathrm{LMD}$, $\mathrm{PFD} = \mathrm{KMD}$,
and P and Q are the two points required.
We now proceed to show how, by combining
these two propositions, the circles capable of simultaneously
touching three given circles may be found.
Let A, B, C, be the centres of the three circles,
and let the sides of the triangle ABC be as usual
denoted by $a$, $b$, $c$; the radii of the circles being
$\alpha$, $\beta$, $\gamma$.
We will suppose that the circle required envelopes
A and touches B and C externally, and the same
process, mutatis mutandis, will give the other
circles.
Taking AB for axis of $x$, and A for origin, we
easily find in the usual way the equation to the
hyperbola, which is the locus of the centres of the
circles touching A and B.
\[r^2-x^2=(r+\alpha+\beta)^2-(c-x)^2\]
\[⇒
r = \frac{c^2-(\alpha + \beta)^2 - 2cx}{2(\alpha + \beta)}
\] From which, DF being the citerior directrix, we
have
\[
\mathrm{AD} = \dfrac{c^2-(\alpha + \beta)^2}{2c}
\]
Hence, with radius
$\mathrm{BK} = \alpha + \beta$ describe an arc. Bisect AB, and
from its middle point as centre and rad. $\tfrac{1}{2} \mathrm{AB}$
describe an arc, intersecting the former in K. Draw
$\mathrm{KN} \perp \mathrm{AB}$, and bisect AN in D, then $\mathrm{DF}
\perp \mathrm{AB}$ is the citerior directrix. Again, make AV
to AD as $c$ to $\alpha + \beta + c$, i.e. as AB to rad. A + rad. B + AB, and V will be the citerior
vertex.
Assign the citerior directrix EF of the hyperbola,
which is the locus of the circles touching
A and C. Make DG to EH in the ratio
compounded of the ratios of $b$ to $c$, and $\alpha + \beta$
to $\alpha + \gamma$. Draw GS and HS $\perp$ to BA and
CA, and through S and F draw SPFQ; this
will be the line of centres, and by applying the
second proposition, two points, P and Q, will be
found. Join PA, and produce it to meet the
circle A in L, and with radius PL describe a circle,
and this will envelope A and touch B and C externally.
Also, if QA be joined, cutting circle A in
$\mathrm{L}'$, and a circle radius $\mathrm{QL}'$ be described, it will
envelope B and C, and touch A externally.
Similarly the three other pairs of circles may be
found.
As it would too much increase the extent of this
work to go seriatim through the several cases of the
tangencies---that is, to apply the foregoing propositions
to each case, the reader is supposed to apply
them himself.
I have in the "Mathematician", vol. I, p. 228,
proposed and proved a curious relation amongst
the radii of the eight tangent circles. The following
is another curious property.
Curious property respecting the directions of hyperbolae which are the loci of centres of circles touching each pair of three circles
With reference to the last figure, suppose we
denote the hyperbolic branch of the locus of the
centres of circles enveloping A and touching B
externally by $\mathrm{A}_c\mathrm{B}_u$, $\mathrm{A}_u\mathrm{B}_c$, the former meaning
"branch citerior to A and ulterior to B", the latter
"citerior to B and ulterior to A". The six hyperbolic
branches will then be thus denoted:
\[
\mathrm{A}_c \mathrm{B}_u ,\;\; \mathrm{A}_u\mathrm{B}_c ;\;\;
\mathrm{B}_c\mathrm{C}_u ,\;\; \mathrm{B}_u \mathrm{C}_c ;\;\;
\mathrm{C}_c \mathrm{A}_u ,\;\; \mathrm{C}_u\mathrm{A}_c
\]
and suppose the corresponding directrices denoted
thus:
\[
\overline{ \mathrm{A}_c\mathrm{B}_u } ,\;\;
\overline{ \mathrm{A}_u\mathrm{B}_c } ;\;\;
\overline{ \mathrm{B}_c\mathrm{C}_u } ,\;\;
\overline{ \mathrm{B}_u\mathrm{C}_c } ;\;\;
\overline{ \mathrm{C}_c\mathrm{A}_u } ,\;\;
\overline{ \mathrm{C}_u\mathrm{A}_c }
\]
Then the point P is the mutual intersection of
\[
\mathrm{A}_c \mathrm{C}_u ,\;
\mathrm{A}_c \mathrm{B}_u ,\;
\mathrm{B}_c \mathrm{C}_u
\]
and Q is the mutual intersection of
\[
\mathrm{A}_u \mathrm{C}_c ,\;
\mathrm{A}_u \mathrm{B}_c ,\;
\mathrm{B}_u \mathrm{C}_c
\]
PQ passes through intersection of $\overline{\mathrm{A}_c \mathrm{B}_u}$,
$\overline{\mathrm{A}_c \mathrm{C}_u}$
because it passes through intersections of $\mathrm{A}_c \mathrm{B}_u$,
$\mathrm{A}_c \mathrm{C}_u$, and of $\mathrm{A}_u \mathrm{C}_c$, $\mathrm{A}_u \mathrm{B}_c$.
Also PQ through $\overline{\mathrm{B}_c \mathrm{A}_u}$,
$\overline{\mathrm{B}_c \mathrm{C}_u}$, because through
$\mathrm{B}_c \mathrm{A}_u$, $\mathrm{B}_u \mathrm{C}_c$ and
$\mathrm{B}_u \mathrm{A}_c$, $\mathrm{B}_c \mathrm{C}_u$.
Also PQ through $\overline{\mathrm{C}_c \mathrm{A}_u}$,
$\overline{\mathrm{C}_c \mathrm{B}_u}$, because through
$\mathrm{C}_c \mathrm{A}_u$, $\mathrm{C}_c \mathrm{B}_u$ and
$\mathrm{C}_u \mathrm{A}_c$, $\mathrm{C}_u \mathrm{B}_c$, and hence the intersections
$\overline{\mathrm{A}_c \mathrm{B}_u}$, $\overline{\mathrm{A}_c \mathrm{C}_u}$;
$\overline{\mathrm{B}_c \mathrm{A}_u}$, $\overline{\mathrm{B}_c \mathrm{C}_u}$;
$\overline{\mathrm{C}_c \mathrm{A}_u}$, $\overline{\mathrm{C}_c \mathrm{B}_u}$ are all in
the same straight line PQ.
That is, the intersections of pairs of directrices
citerior respectively to A, B, C are in the same
straight line, namely, the line of centres of the pair
of tangent circles to which they belong.
本帖最后由 hbghlyj 于 2021-12-24 07:33 编辑 Chapter Ⅱ 本章的主题是通过一些定点并与一些定直线相切的二次曲线.
Let $u=0$, $v=0$, $w=0$, be the equations to three
given straight lines.
The equation
\[
\lambda vw+\mu uw+\nu uv=0 \tag{1}
\]
being of the second order represents a conic section,
and since this equation is satisfied by any two of
the three equations $u=0$, $v=0$, $w=0$, (1) will pass
through the three points formed by the mutual
intersections of those lines.
To assign values of $\lambda$, $\mu$, $\nu$, in terms of the co-ordinates
of the centre of (1),
We have
\begin{align*}
u &= a_2x+b_2y+1 \\
v &= a_3x+b_3y+1 \\
w &= a_4x+b_4y+1
\end{align*}
Hence (1) differentiated relatively to $x$ and $y$ will
give
\[
\begin{aligned}
\lambda\{a_4v+a_3w\}+\mu\{a_2w+a_4u\}+\nu\{a_3u+a_2v\}&=0\\
\lambda\{b_4v+b_3w\}+\mu\{b_2w+b_4u\}+\nu\{b_3u+b_2v\}&=0
\end{aligned}
\tag{a}
\]
and these are the equations for finding the co-ordinates
of the centre.
Let now L, M, and N be three such quantities that
\[
\mathrm{L}u+\mathrm{M}v+\mathrm{N}w
\]
may be identically equal to 2K, then by finding
the ratios $\dfrac{\lambda}{\mu}$, $\dfrac{\lambda}{\nu}$ from (a) it will be found that the
following values may be assigned to $\lambda$, $\mu$, $\nu$,
\[
\lambda = u (Lu - K), \quad \mu = v (Mv - K), \quad \nu = w (Nw - K)
\]
Hence when any relation exists amongst $\lambda$, $\mu$, $\nu$, we
can, by the substitution of these values, immediately
determine the locus of the centres of (1).
Locus of centres of all conic sections through same four points
$1^\textrm{o}$ Let (1) pass through a fourth point, then $\lambda$, $\mu$, $\nu$,
are connected by the relation
\[
\mathrm{A} \lambda + \mathrm{B} \mu + \mathrm{C} \nu = 0 \tag{$\alpha$}
\]
where A, B, C are the values of $vw$, $uw$, $uv$, for
the fourth point.
Hence the locus of the centres of all conic sections
drawn through the four points will be
\[
\mathrm{A} u (\mathrm{L}u - \mathrm{K}) + \mathrm{B} v (\mathrm{M}v -
\mathrm{K}) + \mathrm{C} w (\mathrm{N}w - \mathrm{K}) = 0 \tag{b}
\]
which is itself a curve of the second order.
这条锥线就是Nine-point conic
Locus of centres of all conic sections through two given points, and touching a given line in a given point
$2^\textrm{o}$ When the fourth point coincides with one of
the other points, the values of A, B, C vanish. But
suppose the fourth point infinitely near to the intersection
of $u = 0$, $v = 0$, and that it lies in the straight
line $u + n v = 0$. Then since on putting $x + h$, $y + k$
for $x$ and $y$, we have
\begin{align*}
& & (v w)^1 = &(v w) + (a_4 v + a_3 w) h + (b_4 v + b_3 w) k && \\
&\text{and } \therefore
& &\mathrm{A} = \mathrm{w} (a_3 h + b_3 k) &&\\
& & &\mathrm{B} = \mathrm{w} (a_2 h + b_2 k)&&\\
& & &\mathrm{C} = 0 &&
\end{align*}
Where w is the value of $w$, for the values of $x$ and $y$
determined by $u = 0$, $v = 0$.
Moreover from the equation $u + n v = 0$
\[
a_2 h + b_2 k + n (a_3 h + b_3 k) = 0
\]
\begin{align*}
&\text{Hence} & \mathrm{B} + n \mathrm{A} & = 0 &&\\
& \text{and} & \mathrm{A} \lambda + \mu \mathrm{B} & = 0 &&\\
& \text{and} \therefore & \lambda - n \mu & = 0 &&\\
& \therefore & u (\mathrm{L}u - \mathrm{K})- n v & (\mathrm{M} v - \mathrm{K}) = 0 &&
\end{align*}
is the ultimate state of equation (b). This latter is
therefore the locus of the centres of all conic sections
which can be drawn through two given points
$u w$, $v w$, and touching a given straight line
$u + n v = 0$ in a given point $u v$.
Locus of centres of all conic sections passing through three given points, and touching a given straight line
$3^\textrm{o}$ Let $\lambda$, $\mu$, $\nu$ be connected by the equation
\[
(\mathrm{A} \lambda)^\frac{1}{2} + (\mathrm{B} \mu)^\frac{1}{2} +
(\mathrm{C} \nu)^\frac{1}{2} = 0 \tag{$\beta$}
\]
and in conformity with this condition let us seek
the envelope of (1):
Diff. (1) and ($\beta$) relatively to $\lambda$, $\mu$, $\nu$, we have
\[
vw\;d\lambda + uw\;d\mu + uv\;d\nu = 0
\]
or
\[
\frac{1}{u}\;d \lambda + \frac{1}{v}\;d \mu + \frac{1}{w}\;d \nu = 0
\]\[
\frac{\mathrm{A}^\frac{1}{2}}{\lambda^\frac{1}{2}}\;\,d\lambda +
\frac{\mathrm{B}^\frac{1}{2}}{ \mu^\frac{1}{2}}\;\,d\mu +
\frac{\mathrm{C}^\frac{1}{2}}{ \nu^\frac{1}{2}}\;\,d\nu = 0
\]Hence
$\lambda^\frac{1}{2} = k \mathrm{A}^\frac{1}{2} u$,
$ \mu^\frac{1}{2} = k \mathrm{B}^\frac{1}{2} v$,
$ \nu^\frac{1}{2} = k \mathrm{C}^\frac{1}{2} w$
putting which in ($\beta$) we have
\[
\mathrm{A}u + \mathrm{B}v + \mathrm{C}w = 0
\]
for the envelope required. We may therefore
consider ($\beta$) as the condition that the curve (1)
passing through three given points may also touch
a given straight line $t = 0$, for we have only to determine
A, B, and C, so that
\[
\mathrm{A}u + \mathrm{B}v + \mathrm{C}w = t
\]
identically. Substituting the values of $\lambda$, $\mu$, $\nu$, in ($\beta$) we
have for the locus of the centres of a system of
conic sections passing through three given points
and touching a given straight line,
\[
\{ \mathrm{A}u (\mathrm{L}u - \mathrm{K}) \}^\frac{1}{2}
+ \{ \mathrm{B}v (\mathrm{M}v - \mathrm{K}) \}^\frac{1}{2}
+ \{ \mathrm{C}w (\mathrm{N}w - \mathrm{K}) \}^\frac{1}{2} = 0
\tag{c}
\]
which being rationalized will be found to be of the
fourth order.
Equation to a conic section touching three given straight lines
$4^\textrm{o}$ Let $u = 0$, $v = 0$, $w = 0$ be the equations to
three given straight lines,
\[
(\lambda u )^{\frac{1}{2}} + (\mu v)^{\frac{1}{2}} + (\nu w)^{\frac{1}{2}} = 0 \tag{2}
\]
will be the equation necessary to a conic section
touching each of those lines.
For the equation in a rational form is
\[
\lambda^2 u^2 + \mu^2 v^2 + \nu^2 w^2 = 2 \{\lambda \mu u v + \lambda \nu u w + \mu \nu v w\}
\]
Make $w = 0$ and it reduces to
\[
(\lambda u - \mu v)^2 = 0
\]
and hence the points common to (2) and $w = 0$
will be determined by the simultaneous equations
$w = 0$ and $\lambda u - \mu v = 0$. But these being linear,
determine only one point. Hence $w = 0$ is a tangent
to (5). Similarly $u = 0$, $v = 0$ are tangents.
Equation to a conic section touching four given straight lines
$5^\textrm{o}$ Let $\lambda$, $\mu$, $\nu$, be connected by the equation
\[
\frac{\lambda}{\mathrm{A}}+\frac{\mu}{\mathrm{B}}+\frac{\nu}{\mathrm{C}}=0\tag{$\gamma$}
\]
where A, B, C are fixed constants, and consistently
with this condition let us seek the envelope of (2).
Differentiating (2) and ($\gamma$) with respect to $\lambda$, $\mu$, $\nu$,
\[
\lambda^{-\frac{1}{2}} u^{\frac{1}{2}}\; d\lambda + \mu^{-\frac{1}{2}} v^{\frac{1}{2}}\; d\mu + \nu^{-\frac{1}{2}} w^{\frac{1}{2}}\; d\nu = 0
\]
\[
\frac{d \lambda}{\mathrm{A}}+\frac{d\mu}{\mathrm{B}}+\frac{d\nu}{\mathrm{C}} = 0
\]
\begin{align*}
&\text{Hence }
& & \frac{k}{\mathrm{A}} = \lambda^{-\frac{1}{2}} u^{\frac{1}{2}}
\qquad \text{or} \qquad k^2 \lambda = \mathrm{A}^2u && &&
\end{align*}
$k$ being an arbitrary factor.
\begin{align*}
&\text{Also} & & k^2 \mu=\mathrm{B}^2v \qquad k^2\nu=\mathrm{C}^2w&& &&
\end{align*}
putting which in (2) we have
\[
\mathrm{A}u+\mathrm{B}v+\mathrm{C}w=0
\]
for the envelope required, and which being linear
represents a straight line.
Hence, if $t = 0$ be the equation to a fourth
straight line, and $A$, $B$, $C$ be determined by
making
\[
\mathrm{A}u+\mathrm{B}v+\mathrm{C}w\quad\text{identical with} \quad t
\]
equation (2) subject to the condition ($\gamma$) will represent
all conic sections capable of simultaneously
touching four given straight lines,
\[
t = 0, \quad u = 0, \quad v = 0, \quad w = 0.
\]
Expanding the equation (2) into its rational integral
form, and differentiating with respect to
$x$ and $y$, and putting the differential co-efficients
$\dfrac{d(2)}{dx}$, $\dfrac{d(2)}{dy}$ separately $= 0$, we get two equations
for the co-ordinates of the centre. Those equations
may be exhibited thus:
\[
\frac{\lambda (a_2 b_4 - a_4 b_2) + \mu (a_4 b_3 - a_3 b_4)}{w} \qquad \qquad \qquad \qquad
\]
\[
= \frac{\nu (a_4 b_3 - a_3 b_4) + \lambda (a_3 b_2 - a_2 b_3)}{v}
\]
\[
= \frac{\mu (a_3 b_2 - a_2 b_3) + \nu (a_2 b_4 - a_4 b_2)}{u}
\]
\[
\text{or,} \qquad \frac{\lambda\mathrm{M}+\mu\mathrm{L}}{w}=\frac{\nu\mathrm{L}+
\lambda\mathrm{N}}{v}=\frac{\mu\mathrm{N}+\nu\mathrm{M}}{u}
\]
where L, M, N are determined as before by making
\[
\mathrm{L}u+\mathrm{M}v+\mathrm{N}w=2\mathrm{K}\quad\text{identically.}
\]
The preceding equations give
\begin{align*}
\lambda & =\mathrm{L}(\mathrm{L}u-\mathrm{K}) \\
\mu & =\mathrm{M}(\mathrm{M}v-\mathrm{K}) \\
\nu & =\mathrm{N}(\mathrm{N}w-\mathrm{K})
\end{align*}
and putting these in the condition ($\gamma$) we have
\[
0=\frac{\mathrm{L}(\mathrm{L}u-\mathrm{K})}{\mathrm{A}}+\frac{\mathrm{M}(\mathrm{M}v-\mathrm{K})}{\mathrm{B}}+\frac{\mathrm{N}(\mathrm{N}w-\mathrm{K})}{\mathrm{C}}
\]
for the locus of centres which being linear in
$u$, $v$, $w$, will be linear in $x$ and $y$, and therefore
represents a straight line.
Locus of centres of all conic sections touching four given straight lines
$6^\textrm{o}$. Resuming again the equation (2), and making
$\lambda$, $\mu$, $\nu$, subject to the condition
\[
(\mathrm{A}\lambda)^\frac{1}{2}+(\mathrm{B}\mu)^\frac{1}{2}+(\mathrm{C}\nu)^\frac{1}{2}=0,
\]
which will restrict the curve (2) to pass through a
given point, A, B, C being the values of $u$, $v$, and $w$,
for that point. Putting in the values of $\lambda$, $\mu$, $\nu$, determined
above, we have
\[
\{\mathrm{AL}(\mathrm{L}u-\mathrm{K})\}^\frac{1}{2}+
\{\mathrm{BM(M}v-\mathrm{K})\}^\frac{1}{2}+
\{\mathrm{CN}(\mathrm{N}w-\mathrm{K})\}^\frac{1}{2}=0
\]
for the locus of centres.
Locus of centres of all conic sections touching three given straight lines, and passing through a given point,
and very curious property deduced as a corollary
Hence the locus of the centres of all conic sections
which touch three given straight lines and
pass through a given point is also a conic section.
Cᴏʀ. From the form of the equation this locus
touches the lines $u = \dfrac{\textrm{K}}{\textrm{L}}$, $v = \dfrac{\textrm{K}}{\textrm{M}}$, $w = \dfrac{\textrm{K}}{\textrm{N}}$, which are
parallel to the given lines and at the same distances
from them respectively wherever the given point
may be situated, L, M, N, K, being independent
of A, B, C. In fact, it is easy to demonstrate that
they are the three straight lines joining the points
of bisection of the sides of the triangle formed by
$u$, $v$, $w$, and hence the following theorem.
If a system of conic sections be described to
pass through a given point and to touch the sides
of a given triangle, the locus of their centres will
be another conic section touching the sides of the
co-polar triangle which is formed by the lines joining
the points of bisection of the sides of the
former.
Equation to a conic section touching two given straight lines,
and passing through two given points and locus of centres
$7^\textrm{o}$. We now proceed to the case of a conic section
touching two given straight lines, and passing
through two given points. Let $u = 0$, $v = 0$, be
the equations of the two lines touched, and $w = 0$
the equation of the line passing through the two
given points. Then taking the equation
\[
(\lambda u)^\frac{1}{2} + (\mu v)^\frac{1}{2} + (\nu w + 1)^\frac{1}{2} = 0
\]
we know by the preceding that this represents a
conic section touching $u = 0$, $v = 0$, and $w + \dfrac{1}{\nu}= 0$.
Let $\alpha$, $\alpha'$ be the values of $u$ at the given points, and
$\beta$, $\beta'$ those of $v$, the values of $w$ being zero for each,
then the equations for finding $\lambda$ and $\mu$ will be
\begin{align*}
&(\lambda \alpha)^\frac{1}{2}\; + (\mu \beta)^\frac{1}{2}\; + 1 = 0 \\
&(\lambda \alpha')^\frac{1}{2} + (\mu \beta')^\frac{1}{2} + 1 = 0;
\end{align*}
Let A and B be the values of $\lambda$ and $\mu$ deduced from
these, and we have for the equation of the conic
section
\[
(\mathrm{A}u)^\frac{1}{2}+(\mathrm{B}v)^\frac{1}{2}+(\nu w+1)^\frac{1}{2}=0
\]
in which $\nu$ is the only arbitrary constant.
Differentiating this equation when expanded into
its rational form with respect to $x$ and $y$, we have
two equations respectively equivalent to
\[
\nu=-\mathrm{N}\cdot\frac{\mathrm{A}u-\mathrm{B}v}{\mathrm{L}u-\mathrm{M}v}
\]
\[
\nu w+1=\frac{\mathrm{AM}+\mathrm{BL}}{\mathrm{AM}-\mathrm{BL}}(\mathrm{A}u-\mathrm{B}v)
\]
determining L, M, N, as before, by making
\[
\mathrm{L}u+\mathrm{M}v+\mathrm{N}w=2\mathrm{K}\quad \text{identically.}
\]
Hence eliminating $\nu$, there arises
\[
2 \{\mathrm{B}\mathrm{L}(\mathrm{L}u-\mathrm{K})-\mathrm{A}\mathrm{M}(\mathrm{M}v-
\mathrm{K})\}(\mathrm{A}u-\mathrm{B}v)
=(\mathrm{A}\mathrm{M}-\mathrm{B}\mathrm{L})(\mathrm{L}u-\mathrm{M}v)
\]
for the locus of the centres.
This is also a curve of the second order, and the
values of A and B are
\begin{align*}
&\mathrm{A}=\left\{\frac{\beta^\frac{1}{2}-\beta^{'\frac{1}{2}}}
{(\alpha\beta^{'})^\frac{1}{2}-(\alpha^{'}\beta)^\frac{1}{2}}\right\}^2
&\mathrm{B}=\left\{\frac{\alpha^{'\frac{1}{2}}-\alpha^\frac{1}{2}}
{(\alpha\beta{'})^\frac{1}{2}-(\alpha{'}\beta)^\frac{1}{2}}\right\}^2
\end{align*}
This demonstration assumes that it is possible to
draw a tangent to each of the system of curves
parallel to $w = 0$. But in case the given points are
in opposite vertical angles of the given straight
lines, and the curves therefore hyperbolas, this will
not be possible, and accordingly in such case the
values of A and B become imaginary, for in this
case $\alpha$, $\alpha'$, as also $\beta$, $\beta'$ have different signs. The
following method is free from this and every objection,
and is perfectly general.
Another mode of investigating preceding
$8^\textrm{o}$. Let $u = 0$, and $v = 0$, as before, be the equations
to the tangents, $w = 0$ the straight line joining
the two given points, and $w' = a'_4 x + b'_4 y$, $a'_4$
and $b'_4$ being determined as follows:
\begin{align*}
w'_{\alpha \beta} &= a'_4 \alpha + b'_4 \beta =(u v)^\frac{1}{2}_{\alpha \beta}\\
w'_{\alpha' \beta'} &= a'_4 \alpha' + b'_4 \beta' =(u v)^\frac{1}{2}_{\alpha' \beta'}
\end{align*}
$\alpha \beta$; $\alpha' \beta'$ being co-ordinates of the given points.
\begin{align*}
&\text{Then} & &(w + m w')^2 = m^2 u v &&
\end{align*}
is the equation to the system in which $m$ is arbitrary.
For $u = 0$, or $v = 0$, each give $w + m w' = 0$,
and therefore $u$ and $v$ each touch the curve, and
$w + m w' = 0$ is the equation to the line joining
their points of contact. Again, by the preceding
determination of $w'$, we have for $x = \alpha$, $y = \beta$,
$w = 0$, and $m^2 w'^{2}_{\alpha \beta} = m^2 (u v)_{\alpha \beta}$
and similar for $\alpha' \beta'$, and hence $m$ remains arbitrary.
Differentiating the equation
\[
(w + m w')^2 = m^2 u v
\]
with respect to $x$ and $y$,
\begin{align*}
&& m^2 u v &= (w + m w')^2 && &&\\
&& m^2 (a_2 v + a_3 u) &= 2 (w + m w') (a_4 + m a'_4) && &&\\
&& m^2 (b_2 v + b_3 u) &= 2 (w + m w') (b_4 + m b'_4) && &&\\
&& m^2\{\mathrm{L}u-\mathrm{M}v\} & =2m(w+mw')(a'_4b_4-a_4b'_4) && &&\\
&& m\{\mathrm{L}u-\mathrm{M}v\} & =2\mathrm{Q}(w+mw') && &&\\
&& m^2\{\mathrm{L}'u-\mathrm{M}'v\} & =-2\mathrm{Q}(w+mw') && &&\\
&& \therefore \quad m & =-\frac{\mathrm{L}u-\mathrm{M}v}{\mathrm{L}'u-\mathrm{M}'v} && &&\\
&& \therefore \quad w+mw' & =w-\frac{\mathrm{L}u-\mathrm{M}v}{\mathrm{L}'u-\mathrm{M}'v}w' && &&\\
&& \frac{(\mathrm{L}u-\mathrm{M}v)^2}{\mathrm{L}'u-\mathrm{M}'v}+2\mathrm{Q} &
\left\{w-\frac{\mathrm{L}u-\mathrm{M}v}{\mathrm{L}'u-\mathrm{M}'v}w'\right\} =0; && &&\\
& \text{or,}
& (\mathrm{L}u-\mathrm{M}v)^2+2\mathrm{Q}\{(\mathrm{L}'u- & \mathrm{M}'v)w-(\mathrm{L}u-\mathrm{M}v)w'\}=0, && &&
\end{align*}
which is an equation of the second order.
Now let $u_{\alpha \beta}$, $v_{\alpha \beta}$ be both positive, and
$u_{\alpha' \beta'}$, $v_{\alpha' \beta'}$
both negative, and therefore the given points in opposite
vertical angles of the straight lines $u = 0$ and
$v = 0$. Then $a'_4$ and $b'_4$ will both be real quantities,
and $\therefore$ also Q, L$'$, M$'$, and $w'$. Also if $u_{\alpha \beta}$,
$v_{\alpha \beta}$ have different signs, as also $u_{\alpha' \beta'}$, $v_{\alpha' \beta'}$ then
$a'_4$,$b'_4$, Q, L$'$, M$'$, and $w'$ will be of the form A$\sqrt{-1}$
and the above equation equally real.
$9^\textrm{o}$. We have now discussed the several cases of
the general problem, whose enunciation is as follows:
Of four straight lines and four points let any four
be given, and draw a system of conic sections
passing through the given points, and touching the
given lines, to investigate the locus of the centres.
We have shown that in every case except two
the locus is a conic section. The two exceptions
are, first, when there are three given
points and a given straight line, in which case the
locus is
\[
\{\mathrm{A}u(\mathrm{L}u-\mathrm{K})\}^\frac{1}{2}+\{\mathrm{B}v(\mathrm{M}v-\mathrm{K})\}^\frac{1}{2}+
\{\mathrm{C}w(\mathrm{N}w-\mathrm{K})\}^\frac{1}{2}=0
\]
which, being rationalized, is of the fourth order.
But some doubt may exist as to whether such
equation may not be decomposable into two quadratic
factors, and thus represent two conic sections.
That such cannot hold generally will best appear
from the discussion of a particular case.
The other case of exception is when the data are
four straight lines, the locus then being a straight
line; but since a straight line may be included
amongst the conic sections, we may say that there
is but one case of exception.
The particular case we propose to investigate is
the following.
Investigation of a particular case of conic sections passing through
three given points, and touching a given straight line; locus of
centres a curve of third order, the hyperbolic cissoid
Through one of the angular points of a rhombus
draw a straight line parallel to a diagonal, and
let a system of conic sections be drawn, each touching
the parallel to the diagonal, and also passing
through the three remaining angular points of the
rhombus, to investigate the locus of centres.
Let $\mathrm{AO} = \mathrm{OD} = 1$, $\tan \mathrm{BAO} = \tan \mathrm{CAO} = m$.
Taking the origin at O the equations are
\begin{align*}
&& && &u = y - m (x + 1) = 0 & &\text{for AB} && && \\
&& && &v = y + m (x + 1) = 0 & &\text{for AC} && && \\
&& && &w = x = 0 & &\text{for BC} && && \\
&& && \textrm{A} u + &\textrm{B} v + \textrm{C} w = x - 1 = 0 & &\text{for KD.} && &&
\end{align*}
To determine A, B, C, we have therefore
\begin{align*}
&\textrm{A} \bigl(y - m (x + 1)\bigr) +\textrm{B} \bigl(y + m (x + 1)\bigr) + \textrm{C} x = x - 1 &&&\text{identically;} \\
&\therefore\mathrm{A} + \mathrm{B}=0\\
&- m \textrm{A} + m \textrm{B}+\textrm{C}=1\\
&- m \textrm{A} + m \textrm{B}=-1\\
&\therefore\textrm{C} = 2, \quad \textrm{A} = \frac{1}{2 m}, \quad \textrm{B} = -\frac{1}{2 m}. && && &&
\end{align*}
Also for finding L, M, N
\begin{align*}
\mathrm{L}\biggl(y-m(x+1)\biggr)+&\mathrm{M}\biggl(y+m(x+1)\biggr)+\mathrm{N}x=2\mathrm{K}\\
&\mathrm{L}+\mathrm{M}=0\\
-m\mathrm{L}&+m\mathrm{M}+\mathrm{N}=0\\
-m\mathrm{L}&+m\mathrm{M}=2\mathrm{K}\\
\therefore\quad\mathrm{N}=-2\mathrm{K},\quad&\mathrm{L}=-\frac{\mathrm{K}}{m},\quad\mathrm{M}=\frac{\mathrm{K}}{m}
\end{align*}
by the substitution of which in (c) we have for the
locus of the centres,
\[
\bigl\{u (u + m) \bigr\}^\frac{1}{2} + \bigl\{v (v - m) \bigr\}^\frac{1}{2}
+ \bigl\{4 m^2 w (2 w + 1) \bigr\}^\frac{1}{2} = 0,
\]
or
\[
\biggl\{ \{y - m (x + 1) \} \{y - m x \} \biggr\}^\frac{1}{2} +
\biggl\{ \{y + m (x + 1) \} \{y + m x \} \biggr\}^\frac{1}{2}
+ 2 m \bigl\{x (2 x + 1)\bigr\}^\frac{1}{2} = 0
\]
and this equation rationalized and reduced gives
\[
(2 x - 1) (2 x + 1) y^2 = 4 m^2 x^3 (2 x + 1)
\]
which resolves into the two
\begin{align*}
&& &2x + 1 = 0 &&\\
&\text{and}&
&y^2 = \frac{4 m^2 x^3}{2x - 1}&&
\end{align*}
the first of these, since
\[
\nu = w (\mathrm{N} w - \mathrm{K}) = -\mathrm{K} x (2 x + 1)
\]
requires $\nu = 0$, which would reduce the equation
of the system to $\lambda v w + \mu u w = 0$, or
$w = 0$, $\lambda v + \mu u = 0$, representing only two straight
lines. Hence $2 x + 1 = 0$
must be rejected, and therefore $y^2 = \dfrac{4 m^2 x^3}{2x - 1}$
is the required locus. Now this curve is essentially one of the third
order, and generated from the hyperbola in the same
manner as the cissoid of Diocles is from the circle.
This we proceed to demonstrate. Genesis and tracing of the hyperbolic cissoid
If $x$ be measured in the contrary direction OA,
the equation may be written
\[
y^2=\frac{4m^2x^3}{2x+1}
\]
Let a hyperbola be described of which the semi
axes are, real $=\frac{1}{4}$ , imag. $=\dfrac{m\sqrt{2}}{4}$, and through the vertex
A draw any line NAP. Draw NQ and make
CM = CQ. Also draw the ordinate RM, cutting
NAP in P, then P will be a point in the curve. For let $y=\alpha x$ be equation to NAP, A being
origin, equation to hyperbola $y^2 = m^2 x + 2m^2 x^2$,
$\;\therefore\;$ for intersection N, $\alpha^2 x = m^2 + 2 m^2 x$
\[
x=-\frac{m^2}{2m^2-\alpha^2}; \quad \therefore \quad \mathrm{AQ}=\frac{m^2}{2m^2-\alpha^2}
\]
\begin{align*}
&\text{Also}& \mathrm{BQ } = \mathrm{AP} &= x,
\quad \frac{m^2}{2m^2-\alpha^2} -x = \mathrm{AB} = \frac{1}{2} &&\\
&& &\frac{2m^2}{2m^2-\alpha^2}=2x+1; &&\\
&& \therefore \quad &2 m^2 - \alpha^2 = \frac{2 m^2}{2 x + 1} &&\\
&& &\alpha^2 = \frac{4 m^2 x}{2 x + 1} = \frac{y^2}{x^2} &&\\
&& &\therefore \quad y^2 = \frac{4 m^2 x^3}{2x + 1}; &&
\end{align*}
Hence the locus of P is the curve in question. To find the asymptotes. Taking the equation
\[
y^2 = \frac{4 m^2 x^3}{2x - 1}
\]
\[
y = \pm 2 m x^{\frac{3}{2}} (2 x)^{-\frac{1}{2}} \left\{1 -
\frac{1}{2 x} \right\}^{-\frac{1}{2}}
\]
\[
= \pm \sqrt{2} m x \left\{1 + \frac{1}{4 x} + \frac{1}{2} \cdot
\frac{3}{4} \cdot \frac{1}{4 x^2} \ \ etc. \right\}
\]
so that for $x$ infinite we have
\[
y = \pm \sqrt{2} m \left(x + \frac{1}{4} \right);
\]
Moreover since $x = \frac{1}{2}$ makes $y$ and $\dfrac{dy}{dx}$ infinite,
the equation
\[
2 x - 1 = 0
\]
gives another asymptote. These asymptotes being drawn, the curve will be
found to consist of three distinct branches, as in the
figure.
Equation to a conic section touching three given
straight lines, and also the conic section passing through the mutual intersections
of the straight lines and locus of centres
$10^\textrm{o}$. What we have hitherto exhibited seems to
be far from being the full extent of applicability
of this method of investigation, as the following
will show.
\begin{align*}
&\text{Let} & &\mathrm{A} v w + \mathrm{B} u w + \mathrm{C} u v = 0& \tag{3}
\end{align*}
be a fixed conic section passing through the intersections
of $u = 0$, $v = 0$, $w = 0$. It is required to find
a system of conic sections, each of which shall touch
the lines $u = 0$, $v = 0$, $w = 0$, and also the curve (3). \begin{align*}
&\text{Let} & &(\lambda u)^\frac{1}{2} + (\mu v)^\frac{1}{2} +
(\nu w)^\frac{1}{2} = 0& \tag{4}
\end{align*}
be the equation to any curve of the system. This already touches $u$, $v$, $w$, and if we assume
\[
(\mathrm{A} \lambda)^\frac{1}{3} + (\mathrm{B} \mu)^\frac{1}{3} +
(\mathrm{C} \nu)^\frac{1}{3} = 0 \tag{$\delta$}
\]
and investigate the envelope of (4) we find it to be
no other than the equation (3). Hence the equation (4), in which $\lambda$, $\mu$, $\nu$, are subject
to the condition ($\delta$), represents the required
system. Hence the locus of the centres of the system is
\[
\{\mathrm{A}\mathrm{L}(\mathrm{L}u-\mathrm{K})\}^\frac{1}{3}+\{\mathrm{B}\mathrm{M}(\mathrm{M}v-\mathrm{K})\}^\frac{1}{3}
+\{\mathrm{C}\mathrm{N}(\mathrm{N}w-\mathrm{K})\}^\frac{1}{3}=0.
\]
Equation to a conic section passing through the mutual intersections
of three tangents to another conic section, and also touching the
latter and locus of centres
$11^\textrm{o}$. Now let
\[
(\mathrm{A} u)^\frac{1}{2} + (\mathrm{B} v)^\frac{1}{2} +
(\mathrm{C} w)^\frac{1}{2} = 0 \tag{5}
\]
be a fixed conic section touching $u = 0$, $v = 0$, $w = 0$,
and let it be required to find a system of conic sections,
each passing through the mutual intersections
of $u = 0$, $v = 0$, $w = 0$, and also touching (5).
\begin{align*}
&\text{Let}& &\lambda v w + \mu u w + \nu u v = 0 && \tag{6}
\end{align*}
be the equation to each curve of the system, and
suppose $\lambda$, $\mu$, $\nu$ connected by ($\delta$) as before. On investigating the envelope of (6) we find it
to be no other than (5). Hence (6), subject to
condition ($\delta$), represents the system required. The locus of the centres in this case will be
\[
\{\mathrm{A} u (\mathrm{L} u -\mathrm{K})\}^\frac{1}{3} + \{\mathrm{B} v (\mathrm{M} v - \mathrm{K})\}^\frac{1}{3}
+ \{\mathrm{C} w (\mathrm{N} w - K)\}^\frac{1}{3} = 0.
\] This curve is double the dimensions of that in
the preceding case, and each result assures us that
were we to find the solution of the following,
“To find the locus of the centres of systems of
conic sections, each of which touches four given
conic sections,” we should have an algebraical
curve of very high dimensions, and not in general
resolvable into factors, each representing a curve of
the second order.
Solution to a problem in Mr. Coombe's Smith's prize paper for 1846
I will conclude this chapter by applying my
method to solve a theorem proposed by Mr. Coombe
in his Smith's Prize Paper of the present year. The theorem is, “If a conic section be inscribed
in a quadrilateral, the lines joining the points of
contact of opposite sides, each pass through the
intersection of the diagonals.” Let $u=0$, $v=0$, $w=0$, $t=0$, be the equations
to the sides of the quadrilateral; Then determining A, B, C, by making
\[
\mathrm{A}u+\mathrm{B}v+\mathrm{C}w=t, \qquad \text{identically} \tag{1}
\]
And subjecting $\lambda$, $\mu$, $\nu$, to the condition
\[
\frac{\lambda}{\mathrm{A}}+\frac{\mu}{\mathrm{B}}+\frac{\nu}{\mathrm{C}}=0 \tag{2}
\]
\begin{align*}
&\text{we have} & &(\lambda u)^\frac{1}{2}+(\mu v)^\frac{1}{2}+(\nu w)^\frac{1}{2}=0& \tag{3}
\end{align*}
for the inscribed conic section. But equation (3) may be put in the form
\[
4\mu\nu v w = (\lambda u - \mu v - \nu w)^2
\]
\begin{align*}
&\text{so that} &&\qquad\lambda u - \mu v - \nu w=0 && &&
\end{align*}
is the equation to the line joining the points of
contact of $v$ and $w$. From (1) we have
\[
\mathrm{A}u+\mathrm{B}v\quad\text{identical with}\quad t-\mathrm{C}w,
\]
so that either of these equated to zero will represent
the diagonal DB, and similarly
$\mathrm{A}u+\mathrm{C}w=0$, or $t-\mathrm{B}v=0$ will represent the
diagonal AC. \begin{align*}
&\text{But from}& &\mathrm{A}u+\mathrm{B}v=0& &&\\
&& &\mathrm{A}u+\mathrm{C}w=0 & &&\\
&\text{and} & &\frac{\lambda}{\mathrm{A}} + \frac{\mu}{\mathrm{B}} + \frac{\nu}{\mathrm{C}} = 0& &&
\end{align*} Eliminating A, B, C, we obtain
\[
\lambda u - \mu v - \nu w=0
\] Hence this line passes through the intersection of
the two diagonals. But this has been shown to be
the line joining the points of contact of the opposite
sides $v$, $w$; such line of contact therefore passes
through the intersection of diagonals. Similarly, the
other line of contact also passes through the intersection
of diagonals.
Chapter Ⅲ
我遵循上一章的方法应用解析几何成功地解决了很多非常一般的问题。我还把它扩展到了三维，发现了许多新的性质和关系，也得到了已知定理的非常简洁和优雅的证明。本章旨在举例说明其中的一些。 Equation to a surface of second order, touching three planes in points
situated in a fourth plane
Let $u$, $v$, $w$, $t$ be linear functions of $x$, $y$, $z$, and
let the planes
\[
u=0, \quad v=0, \quad w=0,
\]
be supposed to touch a surface of the second order
in points situated in the plane $t=0$; then the
equation to that surface will be as follows:
\[
\mathrm{A}^2u^2+\mathrm{B}^2v^2+\mathrm{C}^2w^2 - 2\mathrm{AB}uv - 2\mathrm{AC}uw - 2\mathrm{BC}vw \pm t^2 =0.
\] For suppose $u=0$, the equation becomes
\[
(\mathrm{B}v-\mathrm{C}w)^2 \pm t^2 =0.
\] Taking the upper sign this requires
\[
\mathrm{B}v-\mathrm{C}w=0, \; \text{and} \; t=0;
\]
and taking the lower sign
\[
\mathrm{B}v-\mathrm{C}w+t=0,\; \text{and} \; Bv-Cw-t=0.
\] In either case the point determined by
\[
u=0, \quad \mathrm{B}v-\mathrm{C}w=0, \quad \text{and} \quad t=0,
\]
will be a point in the surface to which the plane
$u=0$ is tangential. In the former case it will
touch the surface in this point only, in the latter it
will touch in this point and cut in the straight
lines
\[
\mathrm{B}v-\mathrm{C}w+t=0,\quad \text{and} \quad \mathrm{B}v-\mathrm{C}w-t=0.
\](
This is the case of the hyperboloid of one sheet. If $t$ be
obliterated there is but one line, and the surface becomes a cone
whose vertex is the common intersection of $u$, $v$, $w$.) In the same way $v=0$, and $w=0$ are also tangent
planes. Now $\mathrm{B}v-\mathrm{C}w=0$ represents a plane through
the common intersection of the planes $v=0$,
$w=0$. Similarly $\mathrm{A}u-\mathrm{B}v=0$ represents a
plane through the common intersection of $u=0$,
$v=0$; and $\mathrm{C}w-\mathrm{A}u=0$ one through the intersection
of $w=0$, $u=0$. Whence, since $\mathrm{C}w-\mathrm{A}u=0$ is a consequence
of $\mathrm{A}u-\mathrm{B}v=0$, and $\mathrm{B}v-\mathrm{C}w=0$, we may
assert the following theorem: Theorems deduced from the above
If a surface of the second order be tangential
to three planes, the planes passing through the
mutual intersections of every two of them and the
point of contact of the third tangent plane, will
intersect in the same straight line. Again, let straight lines be drawn from the point
of mutual intersection of $u=0$, $v=0$, $w=0$, one
in each plane, and let two surfaces of the second
order touch the three planes $u=0$, $v=0$, $w=0$,
in points respectively situated in those straight
lines, then the equations to the two surfaces differing
only in the value of $t$, we have at their
points of intersection
\[
t^2-t'^2=0,
\]
\[
\text{or} \quad t-t'=0, \quad t+t'=0.
\] Hence if two surfaces of the second order touch
three planes in such a manner that the lines joining
points of contact on each plane all pass through
the point of common intersection of the three
planes, the surfaces (if they intersect at all) intersect
in one plane, or else in two planes. In M. Chasles' Memoirs on Cones and Spherical
Conics, translated by the Rev. Charles
Graves, F.T.C.D., I find the following remark: “These theorems might also be demonstrated
by algebraic analysis; but this method, which in
general offers so great advantages, loses them all
in this case, since it often requires very tedious
calculations, and exhibits no connexion between
the different propositions; so that it is only useful
in verifying those which are already known, or
whose truth has been otherwise suggested as probable.” All who have read M. Chasles' Memoirs must greatly admire the exquisite ingenuity and generalization
displayed in them; but I think no one
who well understands the use of analysis, and is
capable of applying it to the utmost advantage,
would readily subscribe to the preceding remark. I would unhesitatingly engage to furnish good
analytical demonstrations of all M. Chasles' theorems,
and as the matter is allied to the subject
of this volume, and would furnish perhaps a
happy illustration to this part of it, I have
adopted the suggestion of a scientific friend to
devote this Chapter to the analytical investigation
of some of Chasles' properties. Equation to a surface of second order
expressed by means of the equations to the cyclic and metacyclic planes
Let $u=0$, $v=0$ be two planes through the
origin, $w=0$ another plane, $r^2 = x^2 + y^2 + z^2$. Then
\begin{align*}
r^2 = m u v + n w \tag{1}
\end{align*}
represents a surface of the second order, of which
$u=0$, $v=0$, are cyclic planes, and $w = 0$ may
be called the metacyclic plane. For $u =$ const. reduces equation (1) to that
of a sphere. The plane $u =$ const. necessarily
intersects this sphere in a circle. But the surface
and plane intersect in the same curve as the sphere
and plane; therefore $u =$ const. intersects the
surface in a circle. That is, all planes parallel
to $u = 0$ intersect the surface in circles, or in
other words $u = 0$ is a cyclic plane. Similarly $v = 0$ is a cyclic plane. If the origin be a point on the surface $w$ is homogeneous
in $x$, $y$, $z$, and is a tangent plane to the
surface in the origin. For $w = 0$ reduces (1) to $r^2 = muv$, from
which, on eliminating $z$, we obtain evidently a
homogeneous linear equation in $x$ and $y$, which will
represent either the origin or two straight lines
through the origin. In either case $w =0$ is a
tangent plane.(This is also evident from the consideration, that when
the constant term in the general equation of the second or
any higher order is zero, the linear part of the equation represents
a plane touching the surface represented by the whole
equation, in the origin.) Moreover, $u$, $v$, and $w$ are proportional to the
perpendiculars from a point $x$, $y$, $z$, upon those
planes respectively, and may be taken equal to
such perpendiculars by the introduction of proper
multipliers. Hence if $\phi$, $\phi'$, $\theta$ are the angles which
$r$ makes with $u$, $v$, $w$, we have
\[
\frac{u}{r} = \sin \phi, \quad \frac{v}{r} = \sin \phi',
\quad \frac{w}{r} = \sin \theta,
\]
by the substitution of which in (1) we obtain the
condition
\[
r \{1 - m \sin \phi \sin \phi'\} = n \sin \theta.
\] Hence the following theorem. General theorems of surfaces of second order
in which one of M. Chasles' conical theorems is included
If in any surface of the second order a point
be taken at which a tangent plane is drawn, and
if, moreover, a chord whose length is $r$ be drawn
from the assumed point, and $\theta$ be the angle it
makes with the tangent plane, $\phi$ and $\phi'$, the angles
with the cyclic planes, then
\[
r \{1 - m \sin \phi \sin \phi'\} = n \sin \theta,
\]
where $m$ and $n$ are constant. When $n = 0$, the surface becomes a cone, origin
at vertex, and $\sin \phi \sin \phi' =$ const., which is one of
Chasles' theorems. When the chord is in either cyclic plane,
\[
r = n \sin \theta.
\] Whence if $\delta$ be the diameter of one of the circular
sections through the origin, and $\eta$ the angle
of inclination of its plane to the tangent plane,
\[
\delta = n \sin \eta, \quad \therefore \quad n = \frac{\delta}{\sin \eta}
\]
This therefore determines $n$, which will in general
be different for different points on the surface.
Also $m$ will be the same for all points on the
surface, because if the origin be changed, the axes
remaining parallel to themselves, the terms of
the second order remain the same. Determination of constants
In order therefore to determine $m$, suppose the
origin changed to the extremity of one of the
principal axes of the solid, the greatest or least
(not the mean). Let A be the length of such
axis, $\varepsilon$ the angle it makes with either cyclic plane,
D the diameter of a circular section through the
extremity of such axis, then since
\[
\sin \theta = 1, \quad \mathrm{for} \quad \theta = 90^\circ,
\quad \mathrm{and} \quad n = \frac{\mathrm{D}}{\cos \varepsilon},
\]
we have
\[
\mathrm{A} \{1 - m\sin^2 \varepsilon\} = \frac{\mathrm{D}}{\cos \varepsilon};
\quad \therefore \quad m = \frac{\mathrm{A} - \mathrm{D} \sec \varepsilon}{\mathrm{A} \sin^2 \varepsilon}.
\] The equation of the surface therefore becomes
\[
r^2 = \frac{\mathrm{A}-\mathrm{D}\sec\varepsilon}{\mathrm{A} \sin^2 \varepsilon} u v +
\frac{\delta}{\sin \eta} w.
\] From this the reader will be able to deduce
many singularly beautiful properties of great
generality. I will here instance one or two of them: Curve of intersection of two concentric
surfaces having same cyclic planes
If two intersecting concentric surfaces of the
second order have the same cyclic planes, they
will intersect in a spherical conic or curve such
that the product of the perpendiculars from any
point in it on the cyclic planes is constant.
\begin{align*}
&\text{Let}& &r^2 = m u v + b^2,& &&\\
&& &r^2 = m' u v + b'^2,& &&
\end{align*}
be the two surfaces. Eliminating $u$ $v$, we have
\[
(m' - m) r^2 = m' b^2 - m b'^2,
\]
the equation to a sphere.
\begin{align*}
&\text{Also} & &(b'^2 - b^2) r^2 = (m b'^2 - m' b^2) u v, &&
\end{align*}
the equation to a cone vertex at origin.
\begin{align*}
&\text {Also} & &(m -m') u v = b'^2 - b^2;&& \\
&& \therefore \quad &uv = \frac{b'^2 - b^2}{m - m'} = \textrm{const.,}&&
\end{align*}
but $u$, $v$ are the perpendiculars from a point $x,$ $y$, $z$
on the cyclic planes; this product is therefore
constant for all points on the common intersection
of the two surfaces. Again let $\Delta$ be the diameter of a sphere touching
a surface of the second order in two points,
and intersecting that surface in circular sections,
then $\Delta = \dfrac{\delta}{\sin \eta}$, and the equation of the surface is
\[
r^2 = \frac{\mathrm{A}-\mathrm{D}\sec\varepsilon}{\mathrm{A}\sin^2 \varepsilon} u v + \Delta w,
\]
the origin being in the surface, and $w = 0$ a tangent
plane through the origin. Let $w = z$, and make $z = 0$, then
\[
\mathrm{r}^2 = \frac{\mathrm{A}-\mathrm{D}\sec\varepsilon}{\mathrm{A} \sin^2 \varepsilon}
\mathrm{u} \mathrm{v},
\]
r, u, v being the values of
$r$, $u$, $v$ when $z = 0$. In an hyperboloid of one sheet the product of the
lines of the angles made by either generatrix with the cyclic planes proved to be
constant, and its amount assigned in known quantities
This is a homogeneous equation of the second
order in $x$ and $y$, and will therefore represent either
the point of contact, or one or two straight lines.
When it represents one straight line the surface
is conical; when two it is the hyperboloid of one
sheet, the two straight lines being generatrices.
Hence the product of the sines of the angles made
by either generatrix with the cyclic planes is constant
and equal to
\[
\left(\frac{\mathrm{u}}{\mathrm{r}} \cdot \frac{\mathrm{v}}{\mathrm{r}} \right) =
\frac{\mathrm{A}\sin^2\varepsilon}{\mathrm{A}-\mathrm{D} \sec \varepsilon}.
\] We now proceed to a few of the properties
of Cones. Generation of cones of the second degree,
and their supplementary cones
General principle. Let
\[
\mathrm{A}x+\mathrm{B}y+\mathrm{C}z=0,\;\qquad \frac{x}{\mathrm{A}}=\frac{y}{\mathrm{B}}=\frac{z}{\mathrm{C}},
\]
be a moveable plane and straight line perpendicular
thereto. If the condition to which the motion of
the plane or line be subjected be such that
combined or not with the equation
\[
\mathrm{A}^2+\mathrm{B}^2+\mathrm{C}^2=1
\]
it leads to a homogeneous equation of the second
order in A, B, C, then the plane will envelope
a cone of the second degree, and the line will
generate another cone of the second degree supplementary
to the former.
\begin{align*}
&\text{For let}& &f(\mathrm{A},\mathrm{B},\mathrm{C}) = 0,& &&
\end{align*}
be the homogeneous relation above supposed.
Then since $x$, $y$, $z$, in the equations to the straight
line are proportional to A, B, C, we can replace
the latter by the former in the homogeneous equation,
and thus have $f(x, y, z) = 0$, for the
surface described by the moveable straight line. \begin{align*}
&\text{Now let}& \frac{x^2}{a^2} & +\frac{y^2}{b^2}-\frac{z^2}{c^2}=0,& &&\\
\end{align*}be the equation of the cone so described by the straight line\begin{align*}
&& & \frac{x}{\mathrm{A}}=\frac{y}{\mathrm{B}}=\frac{z}{\mathrm{C}}; & \\[2mm]
&\textrm{Then}
& \frac{\mathrm{A}^2}{a^2} & +\frac{\mathrm{B}^2}{b^2}-\frac{\mathrm{C}^2}{c^2}=0.&
\end{align*}
To find the surface enveloped by
\[
\mathrm{A}x+\mathrm{B}y+\mathrm{C}z=0,
\]
\begin{align*}
&\text{we have}& \frac{\mathrm{A}}{a^2}\; d &\mathrm{A}+\frac{\mathrm{B}}{b^2}\; d\mathrm{B}-
\frac{\mathrm{C}}{c^2}\; d\mathrm{C}=0, & && \\[3mm]
&& x\,d&\mathrm{A}+y\,d\mathrm{B}+z\,d\mathrm{C}=0;& &&\\
&\therefore&
\frac{\mathrm{A}}{a^2}=\lambda x&,\quad
\frac{\mathrm{B}}{b^2}=\lambda y,\quad
-\frac{\mathrm{C}}{c^2}=\lambda z,& && \\[3mm]
&\text{or}&
\mathrm{A}=\lambda a^2x&,\quad
\mathrm{B}=\lambda b^2y,\quad
\mathrm{C}=-\lambda c^2z.& &&
\end{align*} Putting these in
\[
\mathrm{A}x+\mathrm{B}y+\mathrm{C}z=0,
\]
\[
a^2 x^2 + b^2 y^2 - c^2 z^2 = 0,
\]
the equation to the other cone. This, therefore, establishes the general principle,
and consequently when any property is predicated
of such cones, the only thing necessary
to be done to demonstrate it is to find whether
combined or not with the condition
\[
\mathrm{A}^2+\mathrm{B}^2+\mathrm{C}^2=1,
\]
it leads to a homogeneous result of the second
order in A, B, C. Analytical proofs of some of M. Chasles' theorems
For example. “The sum or the difference of
the angles which each focal line makes with a
side of the cone of the second degree is constant;”
or in other words, if a straight line drawn from the
point of intersection of two given straight lines
makes angles with them whose sum or difference
is constant, the moveable line traces the surface
of a cone. Let the equations of the given lines be
\[
\frac{x}{a}=\frac{y}{b}=\frac{z}{c},
\]
\[
\frac{x}{a'}=\frac{y}{b'}=\frac{z}{c'},
\]
and of the moveable line
\[
\frac{x}{\mathrm{A}}=\frac{y}{\mathrm{B}}=\frac{z}{\mathrm{C}},
\]
$\theta$ and $\theta'$ the two angles;
\[
\therefore \qquad \theta \pm \theta' = 2 \; \alpha \; \mathrm{const.};
\]
\[
\therefore \qquad \cos(\theta \pm \theta')=\cos 2\alpha,
\]
\[
\text{or} \qquad \cos^2\theta+\cos^2\theta'-2\cos 2\alpha\cos\theta\cos\theta'=\sin^2 2\alpha;
\]
\[
\text{but}\qquad\cos\theta=\mathrm{A}a\pm\mathrm{B}b+\mathrm{C}c,
\]
\[\qquad\qquad
\cos\theta'=\mathrm{A}a'+\mathrm{B}b'+\mathrm{C}c'.
\] Putting these in the above, and multiplying
$\sin^2 2\alpha$ by $\mathrm{A}^2+\mathrm{B}^2+\mathrm{C}^2 (=1)$, we have
\[
(\mathrm{A}a+\mathrm{B}b+\mathrm{C}c)^2+(\mathrm{A}a'+\mathrm{B}b'+\mathrm{C}c')^2
\]
\[
-2\cos2\alpha(\mathrm{A}a+\mathrm{B}b+\mathrm{C}c)(\mathrm{A}a'+\mathrm{B}b'+\mathrm{C}c')
\]
\[
=\sin^2 2\alpha(\mathrm{A}^2+\mathrm{B}^2+\mathrm{C}^2),
\]
a homogeneous relation of the second order, and
therefore the proposition is true. But besides establishing the truth of the proposition,
we are enabled immediately to find the
equation of the cone so traced, thus: $x$, $y$, $z$, being written for A, B, C in the above
relation, making
\begin{align*}
u & =ax+by+cz,\\
u' & =a'x+b'y+c'z,\\
r^2 & =x^2+y^2+z^2,
\end{align*}
we have for the equation in question
\[
r^2\sin^2 2\alpha=u^2+u'^2-2uu'\cos 2\alpha.
\] The lines perpendicular to the planes $u=0$, $u'=0$,
are called focal lines for the following reason.
Consider $u'$ const. and $=p'$, then $p'$ is the perpendicular
from the origin on the plane $u'-p'=0$,
and $\sqrt{r^2-p'^2}$ is therefore the distance from a
point $x$, $y$, $z$, in the section of the cone made by
$u'-p'=0$ and the point in which $p'$ intersects
that plane. The above equation gives
\[
(r^2-p'^2)\sin^2 2\alpha=(u-p'\cos 2\alpha)^2;
\]
\[
\therefore \quad \sqrt{r^2-p'^2}= \text{ linear function of } x, y, z.
\] This is a property of the focus and the focus
only. Hence $p'$ passes through the foci of all sections
perpendicular to it. Similarly, a line perpendicular
to $u=0$ passes through foci of all
sections parallel to this plane. Again, if the moveable line makes angles with
the fixed line, such that the product of their
cosines is constant, it will trace a cone of the
second order. The notation being as before, we have
\[
\cos\theta\cos\theta'=\mathrm{ const. }=n,
\]
\[
\text{or} \quad (\mathrm{A}a+\mathrm{B}b+\mathrm{C}c)(\mathrm{A}a'+\mathrm{B}b'+\mathrm{C}c')
=n(\mathrm{A}^2+\mathrm{B}^2+\mathrm{C}^2),
\]
a homogeneous equation of the second order. This establishes the proposition and gives for
the equation of the cone $nr^2=uu'$, wherein $u=0$,
$u'=0$, are called cyclic planes for the following
reason. Consider $u'$ constant and $=p'$, then
\[
r^2=\frac{1}{n} \cdot p'u,
\]
which being the equation to a sphere, the sections
parallel to $u'=0$, will be circular. Similarly, sections parallel to $u=0$ are circular. If two cones be supplementary to each other,
the cyclic planes of the one will be perpendicular
to the focal lines of the other. Let the two supplementary cones be denoted
by
\[
\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=0,
\]
\[
a^2x^2+b^2y^2-c^2z^2=0,
\]
in which $a$ is supposed greater than $b$. Eliminate $x^2$ from the first by means of the
equation $r^2=x^2+y^2+z^2$, and it becomes
\[
r^2=a^2 \left\{ \left(\frac{1}{a^2}+\frac{1}{c^2} \right)z^2
- \left(\frac{1}{b^2}-\frac{1}{a^2} \right)y^2\ \right\}.
\]
Hence the two cyclic planes are
\[
\left(\frac{a^2+c^2}{c^2} \right)^{\frac{1}{2}}z
\pm \left(\frac{a^2-b^2}{b^2} \right)^{\frac{1}{2}}y=0,
\]
\begin{align*}
\mathrm{or}\quad u=\frac{b}{a}\left\{\frac{a^2+c^2}{b^2+c^2}\right\}^{\frac{1}{2}}z\;
+\; \frac{c}{a}\left\{\frac{a^2-b^2}{b^2+c^2}\right\}^{\frac{1}{2}}y=0,\\
u'=\frac{b}{a}\left\{\frac{a^2+c^2}{b^2+c^2}\right\}^{\frac{1}{2}}z\;
-\; \frac{c}{a}\left\{\frac{a^2-b^2}{b^2+c^2}\right\}^{\frac{1}{2}}y=0,\\
\end{align*} Also eliminating $x^2$ from the other cone,
\[
r^2=\frac{1}{a^2}\{(a^2+c^2)z^2+(a^2-b^2)y^2\},
\]
but if $\cos2\alpha=\dfrac{b^2-c^2}{b^2+c^2}$, it is easily found that
this last equation may be put in the form
\[
r^2\sin^2 2\alpha=u^2+u'^2-2uu'\cos2\alpha.
\] But we have shown that this is the form when
the perpendiculars to $u=0$, $u'=0$, are focal
lines. Hence the cyclic planes $u=0$, $u'=0$, of
the first cone are at right angles to the focal
lines of the second or supplementary cone. Now, incidentally, we have also proved that $2\alpha$
the sum or difference of angles which any side
of the second cone makes with its focal lines is
independent of $a$. Hence, if a system of cones
has the same vertex and axis, and be such that
sections made by a plane perpendicular to the
axis, all have the same major axis, the sum or
difference of the angles which any side of one
of the cones of the system makes with its focal
lines will be constant, not merely for the same
cone, but also for all the cones of the system,
this sum or difference being $=2\tan^{-1}\dfrac{c}{b}$.
No one will fail to notice here the striking
analogy between spherical and plane conics. Again, let the plane of $x$ $y$ be parallel to one
system of circular sections, and let the vertex of
the cone be the origin, and the line through the
centres of the circles be the axis of $z$, which line
will in general be inclined to plane $x$ $y$. Take the axes of $x$, $y$ perpendicular to each other. The equation of the cone will be
\[
x^2 + y^2 = n^2 z^2.
\] Now taking $z$ constant and $nz=a$, suppose
we have $x^2 + y^2 = a^2$, and whatever property be
proved in plano respecting this circle, there will
necessarily be a corresponding one of the cone.
There will therefore be no greater analytical difficulty
in proving the conical property than that
in plano. For example. “If two tangent planes be drawn
to a cone of the second order such that their traces
on a cyclic plane are always inclined at the same
angle, the intersection of such planes will trace
out another cone of the second order having a
cyclic plane in common with the first cone”. The analytical proof will be as follows. The equations to the two tangent planes are
\begin{align*}
&x \cos(\theta + \alpha) + y \sin (\theta + \alpha) = nz,\\
&x \cos (\theta - \alpha) + y \sin (\theta - \alpha) =nz.
\end{align*} Adding and dividing by $2\cos\alpha$,
\[
x \cos\theta + y\sin\theta = nz\sec\alpha.
\] Subtracting $\qquad \qquad \quad x\sin\theta - y\cos\theta = 0$. Taking sum of squares $x^2+y^2=n^2z^2\sec^2 \alpha$,
which is another cone having $x$ $y$ for a cyclic plane. As another illustration take the following. Let there be two given straight lines which intersect,
and let a plane perpendicular to the line
bisecting the angle between them be drawn, then
if two planes revolve about the given lines such
that their traces on the transversal plane include
a constant angle, the intersections of such planes
will trace out a cone of the second order which
shall have one of its cyclic planes parallel to
the transversal plane. This proposition is in fact tantamount to proving
that if the base and vertical angle of a triangle
be constant, the locus of the vertex is a circle,
and it is from this plane proposition that Chasles
infers the conical one. The following is the analysis. The equations to the revolving planes have the
form
\begin{align*}
(x + mz)\cos(\theta + \alpha) + y\sin (\theta + \alpha) = 0, \\
(x - mz)\cos(\theta - \alpha) + y\sin (\theta - \alpha) = 0.
\end{align*} The elimination of $\theta$ immediately gives
\[
(x^2 + y^2)\sin2\alpha = m^2z^2\sin2\alpha - 2mzy\cos2\alpha,
\]
which is a cone of the second order having circular
sections parallel to $x$ $y$. It surely cannot be said that analysis loses any
of its usual advantages in the cases here adduced.
For my own part I always conclude that when
analysis does seem to lose any of its usual advantages,
the fault is not in the analysis, but in the
want of dexterity and clearness of analytical conception
in the analyst. I am now about to make a remark to which
I think considerable importance is to be attached. Whatever a plane problem may be, we may
always consider it as the result of one or more
relations between two variables or unknown quantities
$x$ and $y$. Put $\dfrac{x}{z}$ for $x$ and $\dfrac{y}{z}$ for $y$, and we are sure
to have the corresponding conical problem. Mode of extending plane problems to conical problems
Thus in the several investigations of Chapter II.
if we conceive throughout $\dfrac{x}{z}$ and $\dfrac{y}{z}$ to be put for
$x$ and $y$, and moreover consider $z$ constant until
the final result is obtained, we shall have conical
properties corresponding to each of the plane properties. Enunciation of conical problems corresponding to
many of the plane problems in Chap. II.
It will be sufficient to enunciate one or two,
as the reader will easily supply the rest. If a system of cones touch the four planes of
a tetrahedral angle, the diameters of the several
individuals of that system conjugate to a given
fixed plane, will all lie in the same plane. If a system of cones pass through the four
edges of a tetrahedral angle, the diameters of the
several individuals of that system conjugate to a
given fixed plane, will trace out another cone of
the second order. If a system of cones pass through two of the
edges of a pentahedral angle and touch the two
opposite sides, the diameters of the several individuals
of that system conjugate to a given fixed
plane will trace out another cone of the second
order. I have shown, therefore, how from any plane
problem a conical one may be deduced, and to
this class of conical problems I would propose
the name of plano-conical problems. There is
an equally extensive class arising from the intersection
of cones and concentric spheres, to which
the term sphero-conical problems might with propriety
be applied. These requiring a different
management the following illustrations are supplied. “If through two fixed intersecting right lines
two rectangular planes be made to revolve, their
intersection will trace out a cone of the second
order passing through the fixed right lines and
having its cyclic planes at right angles to them.” This is another of Chasles' theorems.
Let $\dfrac{x}{\mathrm{A}}=\dfrac{y}{\mathrm{B}}=\dfrac{z}{\mathrm{C}}$
be the equations of the generating line.
Let the fixed lines be in the plane $x$ $z$ inclined
at an angle of $\alpha$ to axis of $z$. By the property of
right-angled spherical triangles, we have $\cos2\alpha$
equal product of cosines of generating line with
fixed lines, or
\[
(\mathrm{C}\cos\alpha+\mathrm{A}\sin\alpha)
(\mathrm{C}\cos\alpha-\mathrm{A}\sin\alpha)=\cos2\alpha,
\]
where $\mathrm{A}^2+\mathrm{B}^2+\mathrm{C}^2=1$. This equation is therefore
\[
\mathrm{A}^2\cos^2\alpha+\mathrm{B}^2\cos2\alpha-\mathrm{C}^2\sin^2\alpha=0,
\]
\[
\text{or} \qquad x^2\cos^2\alpha+y^2\cos2\alpha-z^2\sin^2\alpha=0.
\] This is the equation to a cone, and on making
$y=0$, it reduces to
\[
x^2\cos^2\alpha-z^2\sin^2\alpha=0,
\]
\[
\text{or} \quad x\cos\alpha-z\sin\alpha=0, \quad x\cos\alpha+z\sin\alpha=0,
\]
and, therefore, the surface passes through the two
fixed lines of which these are the equations. Eliminating $y$ by the equation $r^2=x^2+y^2+z^2$,
\[
x^2\cos^2\alpha+(r^2-x^2-y^2)\cos2\alpha-z^2\sin^2\alpha=0,
\]
\[
\text{or} \qquad r^2\cos2\alpha=z^2\cos^2\alpha-x^2\sin^2\alpha,
\]
and therefore the cyclic planes are
\[
z\cos\alpha+x\sin\alpha=0,
\]
\[
z\cos\alpha-x\sin\alpha=0,
\]
which are perpendicular to the given lines. Sphero-conical problems
2. Let a system of cones of the second order
pass through the four edges of a tetrahedral angle,
to find the surface traced by the axes of each
individual of the system, or in other words, required
the locus of the spherical centres of a
system of spherical conics each passing through
four fixed points on the sphere. Let $u$ and $v$ be two homogeneous functions of
$x$, $y$, $z$ of second order, so that $u=0$, $v=0$ may
represent two cones of that order. Suppose them
to intersect in four lines, then $u+\lambda v=0$ will
for different values of $\lambda$ represent all the cones
having the same vertex, and passing through the
same lines, for any point in any of the lines makes
$u=0$, $v=0$ separately, and therefore satisfies
the above equation. Now in order to find the equations for the directions
of the axes we have, first considering $z$ as
a function of $x$ and $y$,
\[
\frac{du}{dx}+\frac{du}{dz}\cdot\frac{dz}{dx}
+\lambda\left\{\frac{dv}{dx}+\frac{dv}{dz}\cdot\frac{dz}{dx}\right\}=0,
\]
\[
\frac{du}{dy}+\frac{du}{dz}\cdot\frac{dz}{dy}+\lambda\left\{\frac{dv}{dy}+
\frac{dv}{dz}\cdot\frac{dz}{dy}\right\}=0.
\] Putting $\quad \dfrac{dz}{dx}=-\dfrac{x}{z} \; , \;
\dfrac{dz}{dy}=-\dfrac{y}{z} \; , \quad$ conditions which
insure the perpendicularity of the straight line
represented by the preceding with its conjugate
plane, and thus making it peculiar to the axes,
and then eliminating $\lambda$, the resulting equation is
\begin{align*}
& x\left\{\frac{du}{dy}\cdot\frac{dv}{dz}-\frac{du}{dz}\cdot\frac{dv}{dy}\right\}\\[2mm]
+ & y\left\{\frac{du}{dz}\cdot\frac{dv}{dx}-\frac{du}{dx}\cdot\frac{dv}{dz}\right\}\\[2mm]
+ & z\left\{\frac{du}{dx}\cdot\frac{dv}{dy}-\frac{du}{dy}\cdot\frac{dv}{dx}\right\}=0.
\end{align*} Now $u$, $v$ being homogeneous and of the second
order, $\dfrac{du}{dx}$, etc. will be homogeneous and of the
first order. The preceding will therefore be homogeneous
and in general of the third order. Hence,
classifying the curves described on a spherical surface
by the orders of the equations of the concentric
cones by whose intersection with the
spherical surface they are produced, it will follow
that the locus of the centres of a system of spherical
conics of the second order passing through four
given points will be a spherical conic in general
of the third order.
