Chapter 8.4, Problem 50PS

Solution

To calculate : An equation for the cross section of the ellipsoid with the given dimensions and the distance between the foci.

The equation for the cross section of the ellipsoid is $\frac{x^2}{{\left(14\right)}^2}+\frac{y^2}{{\left(8\right)}^2}=1$ and the distance between the foci is $22.96\text{ cm}$ .

Given information :

A lithotripter uses shock waves to break apart kidney stones or gallstones inside the body. Shock waves generated at one focus of an ellipsoid (a three-dimensional shape with an elliptical cross section) reflect to the stone positioned at the second focus.

  

Formula used:

The standard equation of an ellipse with horizontal major axis is,

  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ .

Calculation :

Draw the figure as below:

  

Since major axis is horizontal. Standard equation of an ellipse with major axis is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ .

The length of the major axis is $28\text{ cm}$ . So,

  $\begin{array}{c}2a=28\\ a=\frac{28}{2}\\ a=14\end{array}$

The length of the minor axis is $16\text{ cm}$ . So,

  $\begin{array}{l}2b=16\\ b=\frac{16}{2}\\ b=8\end{array}$

Write the equation of ellipse.

  $\frac{x^2}{{\left(14\right)}^2}+\frac{y^2}{{\left(8\right)}^2}=1$

The coordinates of the foci will be: $\left(\pm ae,0\right)$ . So, find the value of eccentricity $\left(e\right)$ .

The eccentricity $\left(e\right)$ of an ellipse is:

  $\begin{array}{c}{b}^2=a^2\left(1-e^2\right)\\ {\left(8\right)}^2={\left(14\right)}^2\left(1-e^2\right)\\ 64=196-196e^2\\ 196e^2=132\\ e^2=0.67\\ e=0.82\end{array}$

So, the coordinates of the foci will be,

  $\begin{array}{c}{F}_1=\left(\left(14\times 0.82\right),0\right)\\ =\left(11.48,0\right)\end{array}$

  $F_2=\left(-11.48,0\right)$

Distance between $F_1\text{ and }{F}_2$ is,

  $\begin{array}{l}=\sqrt{{\left(-11.48-11.48\right)}^2+{\left(0-0\right)}^2}\\ =\sqrt{{\left(-22.96\right)}^2}\\ =22.96\text{ cm}\end{array}$

Therefore, the equation for the cross section of the ellipsoid is $\frac{x^2}{{\left(14\right)}^2}+\frac{y^2}{{\left(8\right)}^2}=1$ and the distance between the foci is $22.96\text{ cm}$ .