622 Geometry and Trigonometry
m 1 +m 2 =y 0
x 0,
m 1 m 2 =p
x 0.
We also know that the angle between the tangents isφ. We distinguish two situations.
First, ifφ = 90 ◦, thenm 1 m 2 =−1. This implies xp 0 =−1, so the locus is the line
x=−p, which is the directrix of the parabola.
Ifφ = 90 ◦, then
tanφ=
m 1 −m 2
1 +m 1 m 2=
m 1 −m 2
1 +xp 0.
We thus have
m 1 +m 2 =y 0
x 0,
m 1 −m 2 =tanφ+p
x 0tanφ.We can computem 1 m 2 by squaring the equations and then subtracting them, and we
obtain
m 1 m 2 =y 02
4 x^20−
(
1 +
p
x 0) 2
tan^2 φ.This must equalxp 0. We obtain the equation of the locus to be
−y^2 +(x+p)^2 tan^2 φ+ 4 px= 0 ,which is a hyperbola. One branch of the hyperbola contains the points from which the
parabola is seen under the angleφ, and one branch contains the points from which the
parabola is seen under an angle equal to the suplement ofφ.
(A. Myller,Geometrie Analitica ̆(Analytical Geometry), 3rd ed., Editura Didactica ̧ ̆si
Pedagogica, Bucharest, 1972) ̆
609.Choose a Cartesian system of coordinates such that the equation of the parabola is
y^2 = 4 px. The coordinates of the three points areTi( 4 pα^2 i, 4 pαi), for appropriately
chosenαi,i= 1 , 2 ,3. Recall that the equation of the tangent to the parabola at a point
(x 0 ,y 0 )isyy 0 = 2 p(x+x 0 ). In our situation the three tangents are given by
2 αiy=x+ 4 pαi^2 ,i= 1 , 2 , 3.IfPijis the intersection oftiandtj, then its coordinates are( 4 pαiαj, 2 p(αi+αj)). The
area of triangleT 1 T 2 T 3 is given by a Vandermonde determinant: