Aichi University of Education 2003: Focus of a Parabola

Problem

The inside of the parabola $${y=x^2}$$ is a mirror. When light travels straight down parallel to the $${y}$$-axis from a point A$${(a,\ 1)}$$ on the line $${y=1}$$ $${(0 < a < 1 )}$$ , let B be the point where it hits the parabola, and let C be the point where the light reflected at B hits the parabola again.
(1) Let $${\theta}$$ be the angle between the tangent to the parabola at point B and the $${y}$$-axis. Express the value of $${\tan \theta}$$ using $${a}$$ .
(2) Show that the line passing through the two points B and C passes through a fixed point independent of $${a}$$ .

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First, let's try drawing a diagram.

If you draw the diagram, you will see that the problem of the "fixed point" in (2) is about the focus. Or rather, this is the meaning of the "focus." However, to be able to draw the diagram, you need to know how light reflects on a curved surface. I have also drawn the normal line in the diagram above, which is the correct diagram for the angle of incidence and the angle of reflection. However, in this problem, it is designed so that you can calculate it even without the normal line. In other words, (1) serves as a guide to (2).
Let's put the calculation aside for a moment and first appreciate the problem.

When you open the link, the same diagram as the one I drew earlier will appear. You can drag the green point A to move it along the green line segment. When you move it, you can see that it indeed passes through a fixed point.

Now, let's outline the calculation.

(1) Let $${\theta}$$ be the angle between the tangent to the parabola at point B and the $${y}$$-axis. Express the value of $${\tan \theta}$$ using $${a}$$ .

The slope of the tangent to the parabola at B is $${2a}$$ when differentiated. If we let $${\alpha}$$ be the angle it makes with the positive direction of the $${x}$$-axis, then $${\tan \alpha=2a}$$ . Given this, the fact that the normal is perpendicular to the tangent, and that $${\alpha+\theta=90^{\circ}}$$ , considering the signs, we get $${\tan \theta=\dfrac{1}{2a}}$$ . (It can be found without the normal, but it is easier to think about it if it is there.)

(2) Show that the line passing through the two points B and C passes through a fixed point independent of $${a}$$ .

There are likely several methods. For example, if you find the equation of the normal line and find the point D, which is the symmetric point of A with respect to the normal line, then D lies on the line BC, so you just need to find the equation passing through B and D. If (1) were not there and only (2) were, that is how you would solve it.
However, if we consider (1) as an introduction, we can think as follows.
If we let $${\beta}$$ be the angle that the line BC makes with the positive direction of the $${x}$$-axis, then $${\beta=\alpha-\theta}$$ .

In (1), $${\tan \theta=\dfrac{1}{2a}}$$ was found, and since $${\tan \alpha=2a}$$ , by using the addition theorem for tangent, it is calculated as $${\tan \beta=a-\dfrac{1}{4a}}$$ .
Therefore, the equation of the line BC becomes $${y=\left(a-\dfrac{1}{4a} \right)x+\dfrac{1}{4}}$$ , and if we consider this as an identity with respect to $${a}$$ , we can say that it passes through the fixed point $${(0, \ \dfrac{1}{4} )}$$ .

The diagram was created using Cinderella (CindyJS).

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