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

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: math.stackexchange.com/questions/1756391/… Jun 16 '20 at 6:53
• 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. Jun 16 '20 at 10:13

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.