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.

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: $$y - y_0 = N (x - x_0).$$

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 \(\theta_1 = \theta_2\).

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.