Properties of a Line Passing Through the Focus of a Parabola📂Geometry
Properties of a Line Passing Through the Focus of a Parabola
Theorem
For the parabola y2=4px, if a line passing through the focus P(p,0) intersects the parabola at two points, respectively A,B, then
PA1+PB1=p1
Proof
Case 1. a=b
The case where the line passing through the focus is x=p.
Since PA=PB=2p,
PA1+PB1=2p1+2p1=p1
Case 2. b=a
Without loss of generality, it is sufficient to prove the case where b>a.
Looking at the segment dropped to the directrix from A,B, one can see that it forms a trapezoid as shown below.
By the definition of a parabola, we have PA=p+a and PB=p+b, and the lengths of each segment are obtained as above. From the properties of the trapezoid, we get the proportional relation
4pa:p+a=4pb:p+b.
Turning it into an equation gives
(p+b)4pa=(p+a)4pb,
which simplifies to
a(p+b)2=b(p+a)2.
Expanding this, we have
ap2+2abp+ab2=bp2+2bap+ba2⟹(b−a)ab=(b−a)p2,
which means p2=ab.
====PA1+PB1p+a1+p+b1p2+(a+b)p+ab2p+a+b2p2+(a+b)p2p+a+bp1.
Therefore, in any case, PA1+PB1=p1 holds.