91Ó°ÊÓ

The first step is to make a drawing of the given situation

Its is given that one person is standing in the point $A$and one in the point $B$. These two points are $2$miles apart and the point $B$is due west of the person at point $A$

Let the fireworks be represented with ordered pair $(x,y)$. This exercise will require the use of the hyperbola properties, so make a sketch of a hyperbola with its foci in the point $A$and$B$and passing through the point$(x,y)$

Since the different of the distance from the point $(x,y)$to $B$and the distance from $(x,y)$to $A$is the constant $1100$feet

The hyperbola like the one drawn in the previous step has the following of its equation

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

where role="math" localid="1647499058435" $2a=1100$, so it follow that

$a=550$

The distance between the two people in points $A$and $B$is $5280$feet

Since the points $A$and $B$are in the foci of the hyperbola which are $2c$feet apart it follows that

$$\begin{gathered} 2c=10560 \\ c=5280 \end{gathered}$$

In every hyperbola the following equation hold

$b^2=c^2-a^2$

So using the fact that you know the value of $a$and $c$

$$\begin{gathered} b^2=c^2-a^2 \\ ={5280}^2-{550}^2 \\ =27878400-302500 \\ =27575900 \end{gathered}$$

Therefore the equation of the hyperbola in this is

$\frac{x^2}{302500}-\frac{y^2}{27575900}=1$

Since the firework happened due north of the person standing in the point $A$, and you know the firework took place on the hyperbola, what you have to do now is to put in the x-coordinate of the point $A$

Since you drew the hyperbola with its center in the origin, the x-coordinate of the right focus coinciding with the point $A$, is

$x=5280$

So, plug in this value and solve for$y$

$$\begin{gathered} \frac{x^2}{302500}-\frac{y^2}{27575900}=1 \\ \frac{{5280}^2}{302500}-\frac{y^2}{27575900}=1 \\ \frac{27878400}{302500}-\frac{y^2}{27575900}=1 \\ \frac{y^2}{27575900}=\frac{27878400}{302500}-1 \\ \frac{y^2}{27575900}=\frac{22789}{25} \\ {y}^2=2513819044 \\ y=Â\pm 50138 \end{gathered}$$

Since the fireworks occurred due north of the point $A$, the $y-$coordinates has to be in the first quadrant, therefore only the positive value is correct and the firework occurred in

$(x,y)=(5280,50138)$

or $50138$feet due north of the person standing in the point$A$