How can I prove that $\alpha=90\unicode{176}$?

Question

I am seeking help with the following math problem and would appreciate any assistance. How can I prove that the angle $α$ is $90\unicode{176}$ in the following picture:

In the picture, we see a vertical beam (Red) hit a parabolic function $f(x)=ax^2+b$ at an arbitrary point $P(c,f(c))$. The beam reflects with the same angle of incidence as the angle of incidence, relative to the tangent (Blue) $f(x)$, and hits point $F$. The gray line goes through the point $F$ and the point where the tangent hits the horizontal line $y = b$.

I think the easiest way is to approach it through geometry. The length of $FP$ and point $F$ to where the tangent crosses the y-axis are the same, which could be an important factor. The fact that the tangent crosses the horizontal line $y = b$ at $\left(\frac{c}{2}, 0\right)$ could also be useful in a solution.

I would greatly appreciate any solutions, links, or methods that can be offered.

Answer 1

Hint: If you have already proved the assertions in the next to the last paragraph then note that in an isosceles triangle the median and the altitude to the third side are the same.

Answer 2

Consider a generic parabola with a vertical axis of symmetry of equation:

$$y=ax^2+bx+c$$ with the focus thus having coordinates

$$F\equiv\left(-\frac{b}{2a}; \frac{1-\Delta}{4a}\right)$$ Remember that as the value of the coefficient a decreases and thus as the magnitude of the parabola increases, the focus moves away from the vertex. To the parabola, the generic line $y=mx+q$ turns out to be tangent to it at the point $P(c,f(c))$. Let us see what happens if we apply the laws of reflection to the point of tangency $P$ when a ray parallel to the axis of symmetry of the parabola is incident to it (in blue in the figure below).

The tangent line (in orange) to the parabola forms an angle equal to $α$ with the $x$-axis. Recall that the angular coefficient of the tangent line to the parabola is equal to the tangent of the angle $α$: $$m=\tan (\alpha)=\frac{\sin(\alpha)}{\cos(\alpha)}$$

The light ray (in light blue) will form a $90°$ angle with the $x$-axis and consequently a $90°-α$ angle with the line tangent to the parabola. Since around the point of tangency, the parabola can be approximated by the line tangent to it, the light ray incident at point $P$ will be reflected according to the rules of reflection with respect to the plane containing the line and perpendicular to the Cartesian plane. It follows that, the light ray will form an angle equal to $α$ with respect to the perpendicular to that plane and will be reflected at the same angle on the other side of the perpendicular. The light ray will now form an angle with the $x$-axis that we will call $β$ equal to:

$$\beta=\alpha+\frac{\pi}2-\alpha+\alpha=\frac{\pi}2+2\alpha$$

Answer 3

I'm going to call the origin O, the $x$- and $y$-intercepts of the tangent $H$ and $K$ respectively, and the point $(c, 0)$ $C$. You say you've already proven that $FP = FK$ and $OH = HC$. The next step is showing that triangles $KOH$ and $HCP$ are congruent, giving $KH = HP$. Finally use this to show triangles $FPH$ and $FHK$ are congruent, and thus $\alpha$ must be a right angle.

(Also for the future, it's best to give all points of interest names.)

Answer 4

First of all note that the $b$ in your function $y=ax^2+b$ just moves the graph up and down along the $y$ axis. Here for simplicity of the figure I suppose $b=0$.

A parabola is a set of points with equal distance from a focus point and a directrice line. In the above figure, focus is $F$ and the directrice is the horizontal line below the $x$ axis. I extended the vertical beam passing through $P$ to meet the directrice at $H$. Let $PB$ be tangent to the parabola and let it meet $FH$ at $A$. We are going to find interesting things about our construction.

Since $P$ is a point on the parabola, $$PF = PH$$ And since vertical beam is reflected by the parabola to pass through $F$, we can see that $$\widehat{APF} = \widehat{APH}$$ Therefore $\triangle{HPF}$ is isosceles and $PA$ is the perpendicular bisector of $FH$.

Moreover, $PH \parallel FB$ and a little angle tracing shows that $\triangle BFP$ is also isosceles and therefore $FA$ is the perpendicular bisector of $PB$.

So, $FH$ and $PB$ are perpendicular bisectors of each other, which shows $PFBH$ is a rhombus!

Now, it remains to show that $A$ is actually on the $x$ axis. This part is left as an exercise for the interested reader.