Chapter A5.2, Problem 26E

To determine

To show: that a ray of light emanating from one focus will be reflected to the other focus.

Answer to Problem 26E

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

Explanation of Solution

Calculation:

We have the general equation of ellipse as:

  $b^2{x}^2+a^2{y}^2=b^2{a}^2$

Differentiating this equation, we have,

  $2b^{2}x+2a^{2}yy\text{'}=0$

This implies that,

  $y\text{'}=-\frac{b^{2}x}{a^{2}y}$

Let $P\left(x_0,y_0\right)$ be any point on the ellipse, therefore,

  $y\text{'}\left(x_0\right)=-\frac{b^2{x}_0}{a^2{y}_0}$

Let $F_1\left(c,0\right)and{\text{ F}}_2\left(c,0\right)$ be the foci.

Then

  $m_{P{F}_1}=\frac{y_0}{x_0-c}and{m}_{P{F}_2}=\frac{y_0}{x_0-c}$

Let $\alpha and\beta$ be the angles between the tangent line and $P{F}_{1}andP{F}_2$ respectively. Then,

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

Similarly, $\tan \beta =\frac{b^2}{c{y}_0}$

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 $\alpha =\beta$ .