Properties of a Line Passing Through the Focus of a Parabola
Theorem
For the parabola $y^2 = 4px$, if a line passing through the focus $P(p,0)$ intersects the parabola at two points, respectively $A, B$, then $$ {{1} \over {\overline{PA}} } + {{1} \over {\overline{PB}} } = {{1} \over {p}} $$
Proof
Case 1. $a=b$
The case where the line passing through the focus is $x = p$.
Since $\overline{PA} = \overline{PB} = 2p$, $$ {{1} \over {\overline{PA}} } + {{1} \over {\overline{PB}} } = {{1} \over {2p}} + {{1} \over {2p}}= {{1} \over {p}} $$
Case 2. $b \ne a$
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 $\overline{PA} = p + a$ and $\overline{PB} = p + b$, and the lengths of each segment are obtained as above. From the properties of the trapezoid, we get the proportional relation $$ \sqrt{4pa} : p+a = \sqrt{4pb} : p+b $$. Turning it into an equation gives $$ (p+b)\sqrt{4pa} = ( p+a ) \sqrt{4pb} $$, which simplifies to $$ a (p+b)^2 = b ( p+a )^2 $$. Expanding this, we have $$ a p^2 + 2 ab p + ab^2 = b p^2 + 2 ba p + ba^2 \implies (b-a) ab = (b-a) p^2 $$, which means $p^2 = ab$. $$ \begin{align*} & {{1} \over {\overline{PA}} } + {{1} \over {\overline{PB}} } \\ =& {{1} \over {p+a}} + {{1} \over {p+b}} \\ =& {{ 2p + a + b } \over {p^2 + (a+b)p + ab}} \\ =& {{ 2p + a + b } \over {2p^2 + (a+b)p}} \\ =& {{1} \over {p}} \end{align*} $$. Therefore, in any case, $\displaystyle {{1} \over {\overline{PA}} } + {{1} \over {\overline{PB}} } = {{1} \over {p}}$ holds.
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