Chapter 9.4, Problem 23PPS

To determine

To find: the equation to represents with the centre at the semi ellipse tunnel.

Answer to Problem 23PPS

The equation with the centre at the semi ellipse tunnel is $\frac{y^2}{73.96}+\frac{x^2}{53.29}=1$

Explanation of Solution

Given information:

The given opening of semi ellipse tunnel is $14.6\text{meters}$ wide and $8.6\text{meters}$ high,

The equation of an ellipse with vertical axis and centre at origin is, $\frac{y^2}{a^2}+\frac{x^2}{b^2}=1\mathrm{....}\left(1.\right)$

The given opening of semi ellipse tunnel is $14.6\text{meters}$ wide and $8.6\text{meters}$ high

Simplify as,

The height of the opening is $a=8.6$

The length of the minor axis is the width of the opening as,

  $\begin{array}{l}b=\frac{14.6}{2}\\ b=7.3\end{array}$

Put the values of $a=8.6,b=7.3$ in the equation (1.),

  $\begin{array}{l}\frac{y^2}{a^2}+\frac{x^2}{b^2}=1\\ \frac{y^2}{{\left(8.6\right)}^2}+\frac{x^2}{{\left(7.3\right)}^2}=1\\ \frac{y^2}{73.96}+\frac{x^2}{53.29}=1\end{array}$

Thus, the equation with the centre at the semi ellipse tunnel is $\frac{y^2}{73.96}+\frac{x^2}{53.29}=1$ .