Chapter 9, Problem 51SGR

To determine

To find: The where should light hit the mirror so that light will be reflected to $\left(0,-5\right)$ .

Answer to Problem 51SGR

The light sourceshould hit the mirror at point $\left(\frac{40-24\sqrt{5}}{5},\frac{45-12\sqrt{5}}{5}\right)$ so that it will be reflected to $\left(0,-5\right)$ .

Explanation of Solution

Given information:

  $\frac{y^2}{9}-\frac{x^2}{16}=1$

Light source $\left(-10,0\right)$

Light will be reflected to $\left(0,-5\right)$ .

Calculation:

Consider a hyperbolic mirror which is modeled by the upper branch of the hyperbola

  $\frac{y^2}{9}-\frac{x^2}{16}=1\mathrm{.........}(1)$

The mirror reflects light rays directed at one focus towards the other focus.

Compare equation (1) $\frac{y^2}{9}-\frac{x^2}{16}=1$ with standard form of a vertical hyperbola with center at origin $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ as shown below:

  $\begin{array}{c}{a}^2=9\\ a=3\end{array}$

  $\begin{array}{c}{b}^2=16\\ b=4\end{array}$

Use relation between a, b and c and compute the value of c as follows,

  $\begin{array}{l}{c}^2=a^2+b^2\\ c^2=9+16\\ c=5\end{array}$

Hence, foci of the given hyperbola will be

  $\begin{array}{c}\left(h,k\pm c\right)\text{ = }\left(0,0\pm 5\right)\\ \text{= }\left(0,-5\right)\\ \text{= }\left(0,5\right)\end{array}$

A light source located at $\left(-10,0\right)$ should be directed to its one focus $\left(0,5\right)$ . So that it will be reflected to other focus $\left(0,-5\right)$ . Hence, the path of light source should be the line joining points $\left(-10,0\right)$ and $\left(0,5\right)$ .

Compute the equation of the path of the light source using equation of line passing through two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ .

  $\begin{array}{c}\left(y-y_1\right)=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\left(x-x_1\right)\\ \left(y-0\right)=\frac{\left(5-0\right)}{\left(0-(-10)\right)}\left(x-(-10)\right)\\ y=\frac{5}{10}\left(x+10\right)\end{array}$

  $y=\frac{1}{2}\left(x+10\right)\mathrm{........}(2)$

Compute the intersection point of hyperbola (1) $\frac{y^2}{9}-\frac{x^2}{16}=1$ and line (2) $y=\frac{1}{2}\left(x+10\right)$

Substitute $y=\frac{1}{2}\left(x+10\right)$ in equation (1) and simplify,

  $\begin{array}{c}\frac{{\left[\frac{1}{2}\left(x+10\right)\right]}^2}{9}-\frac{x^2}{16}=1\\ 16\left[\frac{1}{4}\left(x^2+20x+100\right)\right]-9x^2=144\\ 4x^2+80x+400-9x^2=144\end{array}$

  $-5x^2+80x+256=0$

Use quadratic formula,

  $\begin{array}{l}x=\frac{-80\pm \sqrt{{80}^2-4\times (-5)\times 256}}{2\times (-5)}\\ x=\frac{40\pm 24\sqrt{5}}{5}\end{array}$

For $x=\frac{40+24\sqrt{5}}{5}$ the value of y will be,

  $\begin{array}{l}y=\frac{1}{2}\left[\frac{40+24\sqrt{5}}{5}+10\right]\\ y=\frac{45+12\sqrt{5}}{5}\end{array}$

For $x=\frac{40-24\sqrt{5}}{5}$ the value of y will be,

  $\begin{array}{l}y=\frac{1}{2}\left[\frac{40-24\sqrt{5}}{5}+10\right]\\ y=\frac{45-12\sqrt{5}}{5}\end{array}$

So, intersection points of given hyperbola and the path of light source are $\left(\frac{40+24\sqrt{5}}{5},\frac{45+12\sqrt{5}}{5}\right)$ and $\left(\frac{40-24\sqrt{5}}{5},\frac{45-12\sqrt{5}}{5}\right)$

The point $\left(\frac{40+24\sqrt{5}}{5},\frac{45+12\sqrt{5}}{5}\right)$ will come after the focus $\left(0,5\right)$ . Therefore, the light sourceshould hit the mirror at point $\left(\frac{40-24\sqrt{5}}{5},\frac{45-12\sqrt{5}}{5}\right)$ so that it will be reflected to $\left(0,-5\right)$ .

Therefore, the light sourceshould hit the mirror at point $\left(\frac{40-24\sqrt{5}}{5},\frac{45-12\sqrt{5}}{5}\right)$ so that it will be reflected to $\left(0,-5\right)$ .