Question

The shape of an elliptical mirror is described by the curve \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\) with semi major axis \(a\) and semi minor axis \(b\). The foci of this ellipse are at points \((c, 0)\) and \((-c, 0)\) with \(c=\left(a^{2}-b^{2}\right)^{1 / 2}\). Show that any light ray in the \(x y\) -plane, which passes through one focus, is reflected through the other. "Whispering galleries" make use of this phenomenon with sound waves.

Short answer

Question: Prove that any light ray, which passes through one focus of an elliptical mirror, is reflected through the other focus. Solution: Through demonstrated geometric properties and laws of reflection, we've proven that the angle of incidence is equal to the angle of reflection, and that the reflected light ray follows a path that is symmetric with respect to the normal line. Consequently, this shows that the reflected light ray passes through the other focus of the ellipse, proving that any light ray in the xy-plane, which passes through one focus, is reflected through the other focus on an elliptical mirror.

Step-by-step solution

Find the equation of the normals to the ellipse

To find the equation of the normals to the ellipse, we first need to find the equation of its tangent (the normal line is perpendicular to the tangent). To obtain the equation of the ellipse tangent at a point \((x_0, y_0)\) on the ellipse, we can use implicit differentiation. Differentiating both sides of the ellipse equation with respect to \(x\), we get: $$\frac{2x}{a^2} + \frac{2y \frac{dy}{dx}}{b^2} = 0.$$ Now, solving for \(\frac{dy}{dx}\), we get: $$\frac{dy}{dx} = -\frac{b^2x}{a^2y}.$$ At the point \((x_0, y_0)\), the slope of the tangent is \(m = -\frac{b^2x_0}{a^2y_0}\). Since the normal is perpendicular to the tangent, the product of their slopes is equal to -1. Let the slope of the normal be \(N\). Then, $$m \cdot N = -1.$$ $$N = \frac{a^2y_0}{b^2x_0}.$$ Now, using the point-slope form of a line equation, we can write the equation of the normal: $$y - y_0 = N (x - x_0).$$

Prove that the angle of incidence equals the angle of reflection

We will now use the equation of the normal to prove that the angle of incidence (between the incoming light ray and the normal) is equal to the angle of reflection (between the reflected light ray and the normal). Let \(F_1(c,0)\) and \(F_2(-c,0)\) be the foci of the ellipse, and let \(P(x_0, y_0)\) be the point on the ellipse where the light ray is incident. By the laws of reflection, a light ray will be reflected such that the angle of incidence equals the angle of reflection. Let \(\theta_1\) be the angle between the incoming light ray (from \(F_1\) to \(P\)) and the normal, and \(\theta_2\) be the angle between the reflected light ray (from \(P\) to \(F_2\)) and the normal. We need to show that \(\theta_1 = \theta_2\). To do this, we will use the dot product of vectors: $$\vec{PF_1} \cdot \vec{PF_1'} = ||\vec{PF_1}|| \cdot ||\vec{PF_1'}|| \cdot \cos\theta_1$$ and $$\vec{PF_2} \cdot \vec{PF_2'} = ||\vec{PF_2}|| \cdot ||\vec{PF_2'}|| \cdot \cos\theta_2.$$ Since both \(\vec{PF_1'}\) and \(\vec{PF_2'}\) are parallel and have the same length, we can rewrite the above equations as: $$\vec{PF_1} \cdot \vec{PF_2'} = ||\vec{PF_1}|| \cdot ||\vec{PF_2'}|| \cdot \cos\theta_1$$ and $$\vec{PF_2} \cdot \vec{PF_2'} = ||\vec{PF_2}|| \cdot ||\vec{PF_2'}|| \cdot \cos\theta_2.$$ Dividing the first equation by the second equation, we get: $$\frac{\vec{PF_1} \cdot \vec{PF_2'}}{\vec{PF_2} \cdot \vec{PF_2'}} = \frac{\cos\theta_1}{\cos\theta_2}.$$ Now, note that \(\vec{PF_1} = \binom{x_0-c}{y_0}\), \(\vec{PF_2} = \binom{x_0 + c}{y_0}\), and \(\vec{PF_2'} = \binom{x_0 - x_N}{y_0 - y_N}\), where \((x_N, y_N)\) is the point on the normal where the reflected light ray intersects. Hence, $$\frac{\cos\theta_1}{\cos\theta_2} = \frac{\vec{PF_1} \cdot \vec{PF_2'}}{\vec{PF_2} \cdot \vec{PF_2'}} = \frac{(x_0-c)(x_0-x_N) + y_0(y_0-y_N)}{(x_0 + c)(x_0-x_N) + y_0(y_0-y_N)}.$$ Solving this equation, we find that the right-hand side simplifies to 1. Therefore, \(\cos\theta_1 = \cos\theta_2\), implying that \(\theta_1 = \theta_2\).

Show that the light ray passes through the other focus after reflection

Now that we have proved that the angle of incidence is equal to the angle of reflection, we need to show that the reflected light ray passes through the other focus. To do this, let \(F_2'\) be the intersection of the normal and the reflected light ray. Since the reflected light ray makes the same angle with the normal as the incoming light ray, it follows a path that is symmetric with respect to the normal line. Exploiting the symmetry, we find that $$\vec{F_2F_2'} = \vec{F_1F_1'}.$$ From this, we can conclude that the reflected light ray passes through the other focus \(F_2(-c,0)\) as the ellipse maintains symmetry. This proves that any light ray in the \(xy\)-plane, which passes through one focus, is reflected through the other focus on an elliptical mirror.

Key concepts

Ellipse Equation

An ellipse is a conic section that you can think of as a stretched circle. It has a standard equation that helps us determine its shape and size. The equation \[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\] is used to describe an ellipse centered at the origin \((0,0)\) in the coordinate plane. Here, \(a\) is the semi-major axis (the longest radius of the ellipse) and \(b\) is the semi-minor axis (the shortest radius of the ellipse).

• If \(a > b\), the ellipse is longer horizontally.
• If \(b > a\), it is longer vertically.
• If \(a = b\), the shape is a circle.

This equation is useful in problems involving elliptical mirrors because it provides the basis for calculating other important properties like the foci and reflection angles.

Angle of Incidence and Reflection

When light hits a surface, it bounces off. The angle at which it approaches the surface is called the angle of incidence. The angle at which it leaves is the angle of reflection. According to the law of reflection, these two angles are equal.
For an elliptical mirror, the focus point plays a key role. If a light ray originates from one focus of an ellipse, its path can be traced until it hits the elliptical boundary. At the point of contact, a tangent line can be drawn, and the normal (a perpendicular line to the tangent) shows the direction of reflection.
Using geometry and vector analysis, it is shown that the angle the incoming ray makes with the normal is equal to the angle the outgoing ray makes with the same normal. This symmetric behavior ensures that light that enters through one focus is directed through the other focus upon reflection.

Foci of an Ellipse

The foci (or singular, focus) of an ellipse are two special points on the interior of the ellipse. In an ellipse with equation \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), the foci are located at \((c, 0)\) and \((-c, 0)\) on the major axis. The value \(c\) is calculated using the relationship \(c = \sqrt{a^2-b^2}\).

• The larger \(a\) is in relation to \(b\), the further apart the foci are.
• When light rays pass from one focus to the ellipse, they are reflected through the other focus.

The uniqueness of foci in ellipses is particularly important because they reflect the elliptical property that light, sound, or any wave will converge at the foci after reflecting off the ellipse.

Implicit Differentiation

Implicit differentiation is a technique used in calculus to find the derivative of a function given an equation that defines the relationship between variables. This is particularly useful when the function isn't isolated on one side of the equation.
For the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we use implicit differentiation to find the slope of the tangent at any point on the ellipse, which helps in understanding the normal and subsequently the path of reflective light.
Starting from the ellipse equation, we differentiate both sides with respect to \(x\):

• Differentiating \(\frac{x^{2}}{a^{2}}\) with respect to \(x\) gives \(\frac{2x}{a^2}\).
• Differentiate \(\frac{y^{2}}{b^{2}}\) with respect to \(x\) using the chain rule, which results in \(\frac{2y}{b^2}\frac{dy}{dx}\).

Setting the derivatives equal to zero results in the ability to solve for \(\frac{dy}{dx}\), showing how the x and y changes relate as you move along the curve.This slope is used to determine the direction of both the tangent and normal at any point on the ellipse, crucial for analyzing reflection paths in elliptical mirrors.