383] 
PROBLEMS AND SOLUTIONS. 
563 
in the points K and D (viz., if Z TAK = SAD, the two angles being measured in 
opposite directions from AT, AS respectively); then the line KD meets AS in a 
fixed point B, that is, a point the position of which is independent of the magnitude 
of the equal angles. 
To prove this, take A for the origin, and the bisectors of the angle TAS for 
the axes of x and y: then the equation of the conic is 
ax 2 + 2 hxy + by 2 + 2fy + 2 gx = 0; 
the equation of the tangent at the origin, that is, the line AT, is gx + fy— 0; and 
hence the equation of the line AS is gx —fy — 0. Taking y = ax for the equation 
of the line AK, we have, for the coordinates x 1 , y x of the point K where this meets 
the conic, 
(a + 2ha + ba 2 ) x x + 2 (fa + g) = 0, y x = ax x ; 
and 
then the 
equation of the line AD 
will 
be 
y=- 
coordinates x 2 , 
y 2 of the point 
D where this meets 
the 
(a — 2 ha 
+ ba?) x 2 + 2 
(/«• 
+-9) 
= 0, 
The equation of the line 
KD 
is 
X , 
y 
1 
= 0, 
x x , 
ax x , 
1 
x 2 , 
ax 2 , 
1 
that 
is 
ax 
(x x + x 2 ) + y 
x 2 — 
X x )~ 
- 2ax l 
and for the coordinates of the point B where this meets the line AS, the equation 
whereof is gx —fy = 0, we have 
x [fa (x x + x 2 ) + g (x 2 — xf \ — 2fax x x 2 = 0, 
or, as this may be written, 
But we have 
f aJr y = — £ (a + 2ha + ba"), —- % ^ | (a — 2ha + ba?) ; 
and hence the equation is 
/ 
x (— 2 ha) — 2fa = 0, 
9 
giving x = — j, and thence y — for the coordinates of the point B; and, these 
being independent of a, the lemma is seen to be true. 
71—2