Chapter A5.2, Problem 27E

To determine

To show: that a ray of light directed toward one focus of a hyperbolic mirror, as in the accompanying figure, is reflected toward the other focus.

Answer to Problem 27E

Since $tan\alpha =\tan \beta$ , and $\alpha \text{ and }\beta$ are acute angles, we have $\alpha =\beta$ .

Explanation of Solution

Calculation:

We have,

  $\begin{array}{l}{b}^2{x}^2-a^2{y}^2=b^2{a}^2\\ 2b^{2}x-2a^{2}yy\text{'}=0\\ y\text{'}=-\frac{b^{2}x}{a^{2}y}\end{array}$

Let $P\left(x_0,y_0\right)$ be a point of tangency. The slope from $Pto{F}_1\left(-c,0\right)$ is $\frac{y_0}{x_0+c}$ and from $Pto{F}_2\left(c,0\right)$ it is $\frac{y_0}{x_0+c}$ . Let the tangent through $p$ meet the $x-axis$ in point $A$ , and define the angles $\angle F_{1}PA=\alpha and\angle F_{2}PA=\beta$ . We will now show that $tan\beta =tan\alpha$ .

We have the result,

  $\begin{array}{l}\tan \alpha =\frac{\left(\frac{x_0{b}^2}{y_0{a}^2}-\frac{y_0}{x_0+c}\right)}{1+\left(\frac{x_0{b}^2}{y_0{a}^2}\right)\left(\frac{y_0}{x_0+c}\right)}\\ =\frac{a^2{b}^2+x_0{b}^{2}c}{x_0{y}_0{c}^2+y_0{a}^{2}c}\\ =\frac{b^2}{y_{0}c}\end{array}$

And we have,

  $\begin{array}{l}\tan \beta =\frac{\left(\frac{y_0}{x_0-c}-\frac{x_0{b}^2}{y_0{a}^2}\right)}{1+\left(\frac{x_0{b}^2}{y_0{a}^2}\right)\left(\frac{y_0}{x_0+c}\right)}\\ =\frac{a^2{b}^2+x_0{b}^{2}c}{x_0{y}_0{c}^2-y_0{a}^{2}c}\\ =\frac{b^2}{y_{0}c}\end{array}$

Since $tan\alpha =\tan \beta$ , and $\alpha \text{ and }\beta$ are acute angles, we have $\alpha =\beta$ .

Conclusion:

Since $tan\alpha =\tan \beta$ , and $\alpha \text{ and }\beta$ are acute angles, we have $\alpha =\beta$ .