# Conic sections

1. Nov 1, 2006

I'm hopelessly stuck on this question. Any help will be greatly appreciated.

Prove that if we have a parabolic mirror with focus at F and axis of symmetry the x-axis, then a light ray emmited from F will be reflected parallel to the x-axis.

To prove this consider the parabola y^2=4px (where the focus is at the point (p,0) and the directrix is the line x=-p) and the diagram (below) where N is the point on the directrix which is nearest to P and AP is a tangent to the parabola. Consider the gradients of FN and AP. Use this information to show that the triangle PAF is similar to PAN.

I have managed to show that the gradient of the line PA is (2p/y) and FN is (-y/2p) so those 2 lines are perpendicular. But this is about the only progress I have made with the question.

EDIT:
Sorry, just noticed my diagram is slightly wrong. The point A should be on the y-axis and should also intersect with FN.

Last edited: Nov 1, 2006
2. Nov 1, 2006

### HallsofIvy

Staff Emeritus
Extend line FP beyond the parabola and extend line NP to the right. You should be able to see that, since reflections have angles of incoming and outgoing rays the same, the angle NP extended makes with the tangent (the angle of the incoming ray) is equal to the angle FP (the outgoing ray) makes with the tangent. Then "vertical angles" shows that the angle FP makes with the tangent is the same as is the same as FP extended makes with the tangent- which is the same that the incoming ray, NP extended, makes with the tangent. But then that angel NP extended makes with FP extended is TWICE the angle NP extended makes with the tangent.

And the angle FP makes with NP extended is ("corresponding angles in parallel lines") the same as FP makes with the axis of the parabola: twice the angle NP extended makes with the tangent. Now use the fact that the tangent of that angle is the slope of line FP and that
$$tan(2\theta)= \frac{2tan(\theta)}{1- tan^2(\theta)}$$

3. Nov 2, 2006