Mathematical Immaturity

14.7 Exercises

• Prove that the tangent line at a point $X$ of a parabola bisects the angle between the line joining $X$ to the focus and the line through $X$ parallel to the axis. This gives the reflection property of the parabola. (See Figure 14.3.)
• Let the focus $F = 0,$ then $X - F = X.$ If $\mathbf{u}$ is a unit vector with the same direction as $X,$ then we can define $X$ as $d\mathbf{u}$ where $d = \|X\|.$ Now, let $P$ be a point on the vertical directrix line $L,$ with $X - P$ parallel to the axis of the parabola (i.e., the $x$-axis). By definition, $\|X - P\|$ must be $d.$ In other words, we can rewrite $$ X = P + d\,\mathbf{i}, $$ Taking the derivative of $X$ with respect to $t$ using both definitions, we get $$ X' = d'\,\mathbf{u} + d\,\mathbf{u}'; \quad X' = P' + d'\,\mathbf{i}, $$(since $\mathbf{i}' = 0).$ Taking the dot products with $\mathbf{u}$ and $\mathbf{i},$ respectively, and recalling that $\mathbf{u}'\cdot\mathbf{u} = 0$ and $\mathbf{i}'\cdot\mathbf{i} = 0$ (since $\mathbf{u}$ and $\mathbf{i}$ are of constant length), we get: $$ X'\cdot \mathbf{u} = d'; \quad X'\cdot \mathbf{i} = P'\cdot \mathbf{i} + d'. $$ But $P'$ must be perpendicular to $\mathbf{i}$ since $P$ can only ever move vertically. Thus, $P'\cdot\, \mathbf{i} = 0,$ so that $$ X'\cdot \mathbf{u} = X'\cdot \mathbf{i} = d'. $$ Assuming that the directrix line is to the left of the parabola, this means that the angles between $X$ and $\mathbf{i},$ $X'$ and $\mathbf{i},$ and $X'$ and $\mathbf{u}$ are all such that $0 < \theta < \frac{\pi}{2}.$ Thus, the cosines determined by the above dot products are unique. And since $\mathbf{u} \neq \mathbf{i},$ it follows that the tangent line (with direction $X'$) bisects the angle between the line segment from the focus to $X$ (which has direction $\mathbf{u}$) and the horizontal line through $X$ (which has direction $\mathbf{i}$), thus proving the reflection property of the parabola.