Chapter 8.5, Problem 40PS

Solution

(a)

To find : An equation of the hyperbola with center at the origin that models the June 21 path, given that $a=13.3\text{ meters and }c=28.2\text{ meters}$ .

The equation for the hyperbola with centered at the origin that models the June $21$ path is $\frac{x^2}{176.89}-\frac{y^2}{618.35}=1$ .

Given information :

Each day, except at the fall and spring equinoxes, the tip of the shadow of a vertical pole traces a branch of a hyperbola across the ground. The diagram shows shadow paths for a $20$ meter tall flagpole in Dallas, Texas.

  

Explanation :

Let us consider a hyperbola with the center at the origin with $a=13.3$ meters and $c=28.2$ meters.

To write an equation of the hyperbola, find the value of $b^2$ .

  $\begin{array}{c}{b}^2=c^2-a^2\\ ={\left(28.2\right)}^2-{\left(13.3\right)}^2\\ =795.24-176.89\\ =618.35\end{array}$

Substitute $a=13.3\text{ and }{b}^2=618.35$ into equation of hyperbola with horizontal transverse axis.

  $\begin{array}{c}\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\ \frac{x^2}{{\left(13.3\right)}^2}-\frac{y^2}{618.35}=1\\ \frac{x^2}{176.89}-\frac{y^2}{618.35}=1\end{array}$

Therefore, the equation for the hyperbola with centered at the origin that models the June $21$ path is $\frac{x^2}{176.89}-\frac{y^2}{618.35}=1$ .

(b)

To find : An equation of the hyperbola with center at the origin that models the September 1 path, given that $a=4.1\text{ meters and }c=25.0\text{ meters}$ .

The equation for the hyperbola with centered at the origin that models the September 1 path is $\frac{x^2}{16.81}-\frac{y^2}{608.19}=1$ .

Given information :

Each day, except at the fall and spring equinoxes, the tip of the shadow of a vertical pole traces a branch of a hyperbola across the ground. The diagram shows shadow paths for a $20$ meter tall flagpole in Dallas, Texas.

  

Explanation :

Let us consider a hyperbola with the center at the origin with $a=4.1$ meters and $c=25.0$ meters.

To write an equation of the hyperbola, find the value of $b^2$ .

  $\begin{array}{c}{b}^2=c^2-a^2\\ ={\left(25.0\right)}^2-{\left(4.1\right)}^2\\ =625.00-16.81\\ =608.19\end{array}$

Substitute $a=4.1\text{ and }{b}^2=608.19$ into equation of hyperbola with horizontal transverse axis.

  $\begin{array}{c}\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\ \frac{x^2}{{\left(4.1\right)}^2}-\frac{y^2}{608.19}=1\\ \frac{x^2}{16.81}-\frac{y^2}{608.19}=1\end{array}$

Therefore, the equation for the hyperbola with centered at the origin that models the June $21$ path is $\frac{x^2}{16.81}-\frac{y^2}{608.19}=1$ .