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Properties of a Line Passing Through the Focus of a Parabola 📂Geometry

Properties of a Line Passing Through the Focus of a Parabola

Theorem

20180510\_011438.png

For the parabola y2=4pxy^2 = 4px, if a line passing through the focus P(p,0)P(p,0) intersects the parabola at two points, respectively A,BA, B, then 1PA+1PB=1p {{1} \over {\overline{PA}} } + {{1} \over {\overline{PB}} } = {{1} \over {p}}

Proof

Case 1. a=ba=b

The case where the line passing through the focus is x=px = p.

Since PA=PB=2p\overline{PA} = \overline{PB} = 2p, 1PA+1PB=12p+12p=1p {{1} \over {\overline{PA}} } + {{1} \over {\overline{PB}} } = {{1} \over {2p}} + {{1} \over {2p}}= {{1} \over {p}}


Case 2. bab \ne a

Without loss of generality, it is sufficient to prove the case where b>ab>a.

Looking at the segment dropped to the directrix from A,BA,B, one can see that it forms a trapezoid as shown below.

20180510\_011447.png

By the definition of a parabola, we have PA=p+a\overline{PA} = p + a and PB=p+b\overline{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 \sqrt{4pa} : p+a = \sqrt{4pb} : p+b . Turning it into an equation gives (p+b)4pa=(p+a)4pb (p+b)\sqrt{4pa} = ( p+a ) \sqrt{4pb} , which simplifies to a(p+b)2=b(p+a)2 a (p+b)^2 = b ( p+a )^2 . Expanding this, we have ap2+2abp+ab2=bp2+2bap+ba2    (ba)ab=(ba)p2 a p^2 + 2 ab p + ab^2 = b p^2 + 2 ba p + ba^2 \implies (b-a) ab = (b-a) p^2 , which means p2=abp^2 = ab. 1PA+1PB=1p+a+1p+b=2p+a+bp2+(a+b)p+ab=2p+a+b2p2+(a+b)p=1p \begin{align*} & {{1} \over {\overline{PA}} } + {{1} \over {\overline{PB}} } \\ =& {{1} \over {p+a}} + {{1} \over {p+b}} \\ =& {{ 2p + a + b } \over {p^2 + (a+b)p + ab}} \\ =& {{ 2p + a + b } \over {2p^2 + (a+b)p}} \\ =& {{1} \over {p}} \end{align*} . Therefore, in any case, 1PA+1PB=1p\displaystyle {{1} \over {\overline{PA}} } + {{1} \over {\overline{PB}} } = {{1} \over {p}} holds.