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.

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 ß (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=x/2 and the slope of the normal line is m=-2x. I'm using the following formula to derive tan α and tan ß

tan Φ=(m₁-m₂)/(1+m₁m₂)

However, this formula isn't working for me when m is undefined (i.e., for the vertical line parallel to the axis).

Help very much appreciated!!

Answer 1

I can't give you a complete answer at this time but I can give you some hints.

Let y be the angle between the normal and the vertical. The tan of that angle is 2x.

Draw a figure to see why!

The angle between the focal radius and the normal is made up of 2 parts:

one is the angle the focal radius makes with the x-axis and the other is the complement of y.

Again draw a figure to see why.

Now if you let point p be (a,a²) you can compute a bunch of tan values and you should be able to get a solution for a...I haven't done it yet and it looks messy but possible.

I did look at it with DESMOS and there clearly is a solution.

I will work on the problem some more and get back to you...but maybe another tutor will have given you the answer by then! Good luck.

6/16/2020 10:40 AM

The answer is much simpler than that.

Draw a figure and prolong the vertical until it intersects the directrix

Draw the line segment between this intersection and the focus.

This segment forms the base of an isosceles triangle with base parallel to the normal.

½ of the angle between the focal radius and the normal is equal to the bases angle of this isosceles

triangle (alternate interior angles of parallel lines).

The other half of the angle between the focus radius and the vertical is equal to the other base angle of the isosceles triangle for the same reason.

And there you have it!

I am sorry I cannot reproduce the figure for you, but once you draw it you will be able to follow the logic easily.