Chapter 8.7, Problem 2MRPS

Solution

(a)

To find : An equation for the mirror.

The required equation is $\frac{x^2}{64}-\frac{y^2}{80}=1$ .

Given information :

A hyperbolic mirror reflects light directed toward one focus to the other focus. The hyperbolic mirror shown has foci at $\left(\pm 12,0\right)$ and vertices at $\left(\pm 8,0\right)$ .

  

Explanation :

Consider the figure.

The vertices are $\left(\pm 8,0\right)$ , so

  $a=8$

The foci are at $\left(\pm 12,0\right)$ , so

  $\begin{array}{c}ae=12\\ e=\frac{12}{8}\\ =1.5\end{array}$

Now,

  $\begin{array}{c}{e}^2=1+\frac{b^2}{a^2}\\ 2.25=1+\frac{b^2}{64}\\ 1.25=\frac{b^2}{64}\\ b^2=80\end{array}$

Therefore, the required equation is $\frac{x^2}{64}-\frac{y^2}{80}=1$ .

(b)

To find : An equation for the path of light beam originating at the point $\left(0,9\right)$ before the beam reflects off the mirror.

The required equation is $y+\frac{3}{4}x=9$ .

Given information :

A hyperbolic mirror reflects light directed toward one focus to the other focus. The hyperbolic mirror shown has foci at $\left(\pm 12,0\right)$ and vertices at $\left(\pm 8,0\right)$ .

  

Explanation :

The slope of the line joining $\left(0,9\right)$ and $\left(12,0\right)$ is,

  $\begin{array}{c}m=\frac{0-9}{12-0}\\ =-\frac{3}{4}\end{array}$

So, the equation of the line originating at $\left(0,9\right)$ is,

  $\begin{array}{c}y-9=m\left(x-0\right)\\ y-9=-\frac{3}{4}x\\ y+\frac{3}{4}x=9\end{array}$

Therefore, the required equation is $y+\frac{3}{4}x=9$ .

(c)

To calculate : The point at which light beam originating at the point $\left(0,9\right)$ reflects off the mirror.

The light beam originating at the point $\left(0,9\right)$ reflects off the mirror at $\left(22.32,-7.74\right)$ .

Given information :

A hyperbolic mirror reflects light directed toward one focus to the other focus. The hyperbolic mirror shown has foci at $\left(\pm 12,0\right)$ and vertices at $\left(\pm 8,0\right)$ .

  

Calculation :

The equation for the path of light beam originating at the point $\left(0,9\right)$ before the beam reflects off the mirror is $y+\frac{3}{4}x=9$ .

So,

  $y=9-\frac{3}{4}x$

Substitute the value of $y$ in $\frac{x^2}{64}-\frac{y^2}{80}=1$ .

  $\begin{array}{c}\frac{x^2}{64}-\frac{{\left(9-\frac{3}{4}x\right)}^2}{80}=1\\ \frac{5x^2-4{\left(\frac{36-3x}{4}\right)}^2}{320}=1\\ 5x^2-\frac{1}{4}{\left(36-3x\right)}^2=320\\ 20x^2-1296-9x^2+216x=1280\\ 11x^2+216x-2576=0\end{array}$

Use quadratic formula,

  $\begin{array}{c}x=\frac{-216\pm \sqrt{46656+453376}}{22}\\ =\frac{-216+707.13}{22}\text{ (ignore negative value)}\\ =22.32\end{array}$

Now, substitute the value of $y$ in $y+\frac{3}{4}x=9$ .

  $\begin{array}{c}y+\frac{3}{4}\left(22.32\right)=9\\ y=9-16.74\\ =-7.74\end{array}$

Therefore, the light beam originating at the point $\left(0,9\right)$ reflects off the mirror at $\left(22.32,-7.74\right)$ .