Show that the normal line of a parabola at point P...

Question

Show that the normal line to a parabola at one of its points $P$ bisects the angle included between the focal radius of $P$ and the line through $P$ parallel to the axis of the parabola.

I'm trying to solving this by showing that $\tan α$ (the angle between the line from the focus through $P$ and the normal line) is equal to $\tan \beta$ (the angle between the normal line and the line parallel to the axis through point $P$). I've taken the parabola $4y=x^2$ so that the focus is $(0,1)$, the slope of the tangent line is $y'=m=\frac{x}{2}$ and the slope of the normal line is $m=-2x$. I'm using the following formula to derive $\tan \alpha$ and $\tan \beta$: $$\tan \phi=\frac{m_1-m_2}{1+m_1 m_2}$$ However, this formula isn't working for me when m is undefined (i.e., for the vertical line parallel to the axis).

There is a very simple synthetic proof for that: https://math.stackexchange.com/questions/1756391/tangent-on-parabola-bisects-angle — Intelligenti pauca, Jun 16 '20 at 06:53
Construct on the parallel to the axis a point $Q$ such that $PQ=PF$. If $M$ is the midpoint of $FQ$, then line $PM$ is the angle bisector you need. — Intelligenti pauca, Jun 16 '20 at 10:13

score 1 · Answer 1 · answered Jun 16 '20 at 03:30

Let us consider the parabola $y^2=4ax$ (or $x^2=4ay$ with $a=1$ as you have done, it's only a matter of convenience), with the point $S=(a,0)$ as focus and $O=(0,0)$ the vertex of the parabola.

Let the tangent at point $P$ (whose co-ordinate has parametric form $(at^2,2at)$) intersect the axis of the parabola, the X-axis in our case, at the point $T$.

Let the line through $P$ parallel to the axis, intersect the tangent at $O$ (i.e. the Y-axis) at $T'$.

The line containing $P,T$ has equation $yt=x+at^2$ $\implies$ $T$ has the co-ordinates $(-at^2,0)$.

Then $\mid ST \mid = a+at^2$ which is also the length of the segment $SP$ (which you can find out from their co-ordinates).

Thus the $\Delta SPT$ is isosceles, giving $\angle SPT = \angle STP = \angle TPT'$ where the last equality holds from alternate angles under parallelism.

Thus we have shown that the tangent at $P$, bisects the angle between the focal radius at $P$ and the line through $P$ parallel to the axis of the parabola, and if the tangent at $P$ bisects this angle ($\angle T'PS$ in this case), then the normal will bisect the supplementary angle, which was asked.

Note that this will hold for any parabola, oblique ones as well, because it is only a matter of reparametrisation.

Show that the normal line of a parabola at point P...

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