CONIC SE 
greater than C (23.)j which are equally 
absurd. Tlierefore PH* and QR X QS 
are equal to C D*. 
Cor. 1 . Ifa tangent of a hyperbola meet 
both asymptotes, as P M, the segments 
P H and H M between the asymptotes and 
the point of contact are equal. And if a 
right line cut a hyperbola, or opposite hy- 
perbolas, and both the asymptotes, as Q T, 
the segments between the curve or cuiwes, 
and the asymptotes are equal to one ano- 
ther : that is Cl R = T S, and Q S = T R. 
ForP H% H M*, Q R X Cl S, and T R X 
T S, are all equal to C D*. 
Cor. 2. On the contrary, if P M, inter- 
cepted between the asymptotes, meet the 
hyperbola, in H, and is bisected therej tlien 
P M is a tangent of the hyperbola. 
Cor. 3. If any number of lines all parallel 
to one another, as Cl T and M N , cut a hy- 
perbola and the asymptotes, the rectangles 
Q R X RT, M L X LN, under the seg- 
ments between the curve and the asymp- 
totes are all equal. 
For it is plain (Cor. 1.) that the rectangles 
are all equal to C D*. 
Def. 17. A diameter of a conic section 
that cuts its ordinates at right angles is cal- 
led an axis. 
Cor. Because two conjugate diameters of 
an ellipse, and opposite hyperbolas, cut 
their ordinates in the same angle.', Pr. 18 ; 
therefore, if there be one axis of these 
curves, there will necessarily be two, and 
these will be conjugate diameters, and they 
^ill cut one another at right angles, 
PROP. XXV. 
Fig. 39, 40. An ellipse, and opposite hy- 
perbolas, have two axes. 
Find the centre of the ellipse C (fig. 39.), 
and draw two diameters. 
Then if the two diameters be equal to 
one another, asEF and D I, two other 
diameters, AB andGH, drawn to bisect 
tlie angles contained by D I and E F, will 
be axes of the ellipse. Join E D and D F : 
then the lines A B and Cl H which bisect 
the vertical angles of the isosceles triangles 
FCD and ECD, will bisect the bases 
D E and D F, and likewise cut these lines 
at richt angles. Hence it is plain that A B 
and "g H are conjugate diameters and axes 
of the ellipse. 
But if the two diameters be not equal, as 
M N and P Q, describe a circle from the 
centre C with a radius less than the greater 
semidiameter C M, but greater than the less 
semidiameter CP: then the circle will cut 
the diameter M N on both sides of the, 
centre within the ellipse, and it will be 
without the ellipse towards the point P : 
therefore the circle will cut the periphery 
of the ellipse both between P and M, and 
between P and N : let E and D be the 
points of section ; then two diameters drawn 
through them will be equal, and the axes of 
the ellipse will be found as above. 
In the case of opposite hyperbolas, (fig. . 
39.) find the centi-e C, and from C as a cen- 
tre describe a circle through a point within 
one of the hyperbolas : then that circle will 
cut the hyperbola in two points D and E,. 
and two transverse diameters drawn tlrrough. 
these will be equal to one another; and 
two diameters A B and G H drawn to bisect 
the angles comprehended by the equal dia- 
meters D I and E F will be conjugate dia- 
meters and axes of the hyperbolas. The 
demonstration is the same as for the ellipse. 
PROP. XXVI. 
Fig. 41 and 42. The two axes of an el- 
lipse are always unequal ; and the greater 
axis is the greatest diameter, and the less 
axis the least diameter of the curve. And 
that axis of a hyperbola, which is a trans- 
verse diameter, is the least of all the trans- 
verse diameters. 
Let AB and D E (Fig. 41) be the two 
axes of an ellipse, C the centre, and C H 
any semidiameter ; draw H P perpendicular 
to AB, and HQ perpendicular to DE. 
Because A B and D E are conjugate dia- 
meters ; and H P an ordinate to A B, and 
H Q an ordinate to D E ; therefore, 
AB* : DE* :: AP X PB ; HP*, Cor. .3, 
Def. 15. 
Now, if A B be supposed to be equal to 
D E, it will follow that AP X P B =: H P* ; 
therefore, A P X P B -j- C P* = H 1’* -]- 
C P*, or A C* = C H*. Therefore, A C = 
CH : and the ellipse will be a circle, which 
is not the case. Cor. 9. Therefore, A B 
and D E are unequal; let A B be supposed 
to be greater than D E. 
Because AB* is greater than D E*, there- 
fore A P X P B is greater than H P* ; and 
A P X P B 4- C P*, or A C*, is greater than 
jj pi-j_PC*, or C H*. Therefore the semi- 
axis A C is greater than any othei- semi- 
diameter H C. 
In like manner. 
D E* : A B* D Q X Q E : H Q*. 
Therefore D Q X Q E is less than H Q* ; 
and D Q X Q E + C Q*, or C D*, is le-ss 
tlian H Q* C Q*, or C H*. Therefore the 
