4.1 Geometry 213through this viewpoint of one plane to another are called projective transformations. Up
to a projective transformation there is only one nondegenerate conic—the circle. Any
projectively invariant property that can be proved for the circle is true for any conic
(and by passing to the limit, even for degenerate conics). Such is the case with Pascal’s
theorem: The opposite sides of a hexagon inscribed in a conic meet at three collinear
points. Note that when the conic degenerates into two parallel lines, this becomes Pappus’
theorem.
To conclude our discussion, let us recall that the equation of the tangent line to a
conic at a point(x 0 ,y 0 )is obtained by replacing in the general equation of the conicx^2
andy^2 byxx 0 , respectively,yy 0 ,xybyxy^0 + 2 yx^0 , andxandyin the linear terms byx+ 2 x^0 ,
respectively,y+ 2 y^0.
We now proceed with an example from A. Myller’sAnalytical Geometry(3rd ed.,
Editura Didactic ̆a ̧si Pedagogic ̆a, Bucharest, 1972).
Example.Find the locus of the centers of the equilateral triangles inscribed in the parabola
y^2 = 4 px.
Solution.Let us determine first some algebraic conditions that the coordinates(xi,yi),
i = 1 , 2 ,3, of the vertices of a triangle should satisfy in order for the triangle to be
equilateral. The equation of the median from(x 3 ,y 3 )is
y−y 3
x−x 3=
y 1 +y 2 − 2 y 3
x 1 +x 2 − 2 x 3.
Requiring the median to be orthogonal to the side yields
y 1 +y 2 − 2 y 3
x 1 +x 2 − 2 x 3·
y 2 −y 1
x 2 −x 1=− 1 ,
or
(x 1 −x 2 )(x 1 +x 2 − 2 x 3 )+(y 1 −y 2 )(y 1 +y 2 − 2 y 3 )= 0.So this relation along with the two obtained by circular permutations of the indices are
necessary and sufficient conditions for the triangle to be equilateral. Of course, the third
condition is redundant. In the case of three points on the parabola, namely(y
(^2) i
4 p,yi),
i= 1 , 2 ,3, after dividing byy 1 −y 2 , respectively, byy 2 −y 3 (which are both nonzero),
we obtain
(y 1 +y 2 )(y 12 +y^22 − 2 y 32 )+ 16 p^2 (y 1 +y 2 − 2 y 3 )= 0 ,
(y 2 +y 3 )(y 22 +y^23 − 2 y 12 )+ 16 p^2 (y 2 +y 3 − 2 y 1 )= 0.
Subtracting the two gives