Geometry and Trigonometry 621607.Since the property we are trying to prove is invariant under affine changes of coor-
dinates, we can assume that the equation of the hyperbola is
xy= 1.The asymptotes are the coordinate axes. In the two-intercept form, the equation of the
line is
x
a+
y
b= 1.
Then the coordinates ofAandBare, respectively,(a, 0 )and( 0 ,b). To find the coor-
dinates ofPandQ, substitutey=^1 xin the equation of the line. This gives rise to the
quadratic equation
x^2 −ax+
a
b= 0.
The rootsx 1 andx 2 of this equation satisfyx 1 +x 2 =a. Similarly, substitutingx=^1 yin
the same equation yields
y^2 −by+
b
a= 0 ,
and the two rootsy 1 andy 2 satisfyy 1 +y 2 =b. The coordinates ofPandQare,
respectively,(x 1 ,y 1 )and(x 2 ,y 2 ). We have
AP^2 =(x 1 −a)^2 +y 12 =(a−x 2 −a)^2 +(b−y 2 )^2 =x 22 +(b−y 2 )^2 =BQ^2.The property is proved.
(L.C. Larson,Problem Solving through Problems, Springer-Verlag, 1983)
608.The condition that a line through(x 0 ,y 0 )be tangent to the parabola is that the system
y^2 = 4 px,
y−y 0 =m(x−x 0 )have a unique solution. This means that the discriminant of the quadratic equation inx
obtained by eliminatingy,(mx−mx 0 +y 0 )^2 − 4 px=0, is equal to zero. This translates
into the condition
m^2 x 0 −my 0 +p= 0.The slopesmof the two tangents are therefore the solutions to this quadratic equation.
They satisfy