Question

A tunnel through a mountain for a four-lane highway is to have an elliptical opening.

The total width of the highway (not the opening) is to be $16\mathrm{m}$, and the height at the edge of the road must be sufficient for a truck $4\mathrm{m}$ high to clear if the highest point of the opening is to be $5\mathrm{m}$ approximately.

How wide must the opening be,

Answer 1

Calculate the width of the opening of the tunnel.

A diagram is shown below which describes the situation of the question:

in the diagram, $A\text{'}A$ represent the width of the opening of the tunnel, $B\text{'}B$ represents the width of the highway.

Given that, $B\text{'}B=16\mathrm{m}\Rightarrow OB=8\mathrm{m}$

Since the shape of the tunnel is like an ellipse.

The equation of an ellipse is given as,

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

Where $2a$ represent the length of the major axis and $2b$ represent the length of the minor axis.

Given that, the highest point of the opening is to be $5\mathrm{m}$ approximately.
So, this $5\mathrm{m}$ will be the total height which is the semi-minor axis of the ellipse.

$\therefore b=5\mathrm{m}$

From the figure, it is observed that the point $(8,4)$ lies on the ellipse.

$$\begin{gathered} \therefore \frac{8^2}{a^2}+\frac{4^2}{5^2}=1 \\ \Rightarrow \frac{64}{a^2}+\frac{16}{25}=1 \\ \Rightarrow \frac{64}{a^2}=1-\frac{16}{25} \\ \Rightarrow \frac{64}{a^2}=\frac{9}{25} \\ \Rightarrow a^2=\frac{25\times 64}{9} \\ \Rightarrow a^2=\frac{1600}{9} \\ \Rightarrow a=\sqrt{\frac{1600}{9}} \\ \Rightarrow a=\frac{40}{3} \\ \Rightarrow a=13.34\mathrm{m} \end{gathered}$$

The width of the opening of the tunnel is computed as follow:

$$\begin{gathered} \therefore A\text{'}A=2a \\ =2\times 13.34 \\ =26.68\mathrm{m} \end{gathered}$$

Hence, the width of the opening of the tunnel is $26.68\mathrm{m}$.