214 4 Geometry and Trigonometry
y 13 −y 33 +(y 1 −y 3 )(y^22 − 2 y 1 y 3 )+ 48 p^2 (y 1 −y 3 )= 0.Divide this byy 1 −y 3 =0 to transform it into
y 12 +y 22 +y 32 + 3 (y 1 y 2 +y 2 y 3 +y 3 y 1 )+ 48 p^2 = 0.This is the condition satisfied by they-coordinates of the vertices of the triangle. Keeping
in mind that the coordinates of the center of the triangle are
x=y^21 +y 22 +y 32
12 p
,y=y 1 +y 2 +y 2
3,
we rewrite the relation as
−1
2
(y 12 +y 22 +y 32 )+3
2
(y 1 +y 2 +y 3 )^2 + 48 p^2 = 0 ,then substitute 12px=y 12 +y 22 +y 32 and 3y=y 1 +y 2 +y 3 to obtain the equation of
the locus
− 6 px+27
2
y^2 + 48 p^2 = 0 ,or
y^2 =4 p
9
(x− 8 p).This is a parabola with vertex at( 8 p, 0 )and focus at((^19 +^8 )p,^0 ).
The second problem was given at the 1977 Soviet Union University Student Mathe-
matical Olympiad.
Example.LetP be a point on the hyperbolaxy =4, andQa point on the ellipse
x^2 + 4 y^2 =4. Prove that the distance fromPtoQis greater than 1.
Solution.We will separate the conics by two parallel lines at a distance greater than 1.
For symmetry reasons, it is natural to try the tangent to the hyperbola at the point( 2 , 2 ).
This line has the equationy= 4 −x.
Let us determine the point in the first quadrant where the tangent to the ellipse has
slope−1. If(x 0 ,y 0 )is a point on the ellipse, then the equation of the tangent atxis
xx 0 + 4 yy 0 =4. Its slope is−x 0 / 4 y 0. Setting−x 0 / 4 y 0 =−1 andx 02 + 4 y 02 =4,
we obtainx 0 = 4 /
√
5 andy 0 = 1 /√
- Consequently, the tangent to the ellipse is
y=
√
5 −x.
The distance between the linesy= 4 −xandy=√
5 −xis equal to( 4 −√
5 )/
√
2,
which is greater than 1. Hence the distance between the arbitrary pointsPandQis also
greater than 1, and we are done.