Exercise 10.3.6

Proof. Suppose that $l$ reflects ${\alpha }_1$ to ${\beta }_1\in l_1$ and ${\alpha }_2$ to ${\beta }_2\in l_2$.

Let $d=d(l_1,l_2)$ the distance between $l_1$ and $l_2$. If $\gamma$ is the orthogonal projection of ${\beta }_1$ on $l_2$, then by definition, $d=|{\beta }_1-\gamma |$. By Pythagoras Theorem,

Therefore $d^2=|{\beta }_1-\gamma {\text{|}}^2\le |{\beta }_1-{\beta }_2{\text{|}}^2$, so

Moreover the the reflexion about $l$ preserves the distances between points, so

To conclude, the distance $d$ between $l_1$ and $l_2$ is at most the distance between ${\alpha }_1$ and ${\alpha }_2$.

For instance, let $l_1$ be the line with equation $y=0$, $l_2$ with equation $y=-2$, ${\alpha }_1=i,{\alpha }_2=1+i$. Then $d=d(l_1,l_2)=2>|{\alpha }_1-{\alpha }_2|$, so there is no line $l$ that reflects ${\alpha }_1$ on a point on $l_1$ and ${\alpha }_2$ on a point on $l_2$. In other words, there is no common tangent to the two parabolas with focus ${\alpha }_i$ and directrix $l_i$, $i=1,2$, with equations

(see figure).