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).


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 $T'$.

enter image description here

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.