Chapter 8.7, Problem 43PS

Solution

a.

To find: To find the ship’s position.

The ship is at the point $\left(\text{1, 1}\right)$ .

Given information: The two hyperbolic equations are ${\text{x}}^{\text{2}}{\text{- y}}^{\text{2}}\text{ - 8x + 8 = 0}$ and ${\text{y}}^{\text{2}}{\text{- x}}^{\text{2}}\text{ - 8y + 8 = 0}$ .

Calculation:

The hyperbolic equations are

  $\begin{array}{c}{\text{x}}^{\text{2}}{\text{- y}}^{\text{2}}\text{ - 8x + 8 = 0}\\ {\text{y}}^{\text{2}}{\text{- x}}^{\text{2}}\text{ - 8y + 8 = 0}\end{array}$

Adding both the equations to get,

  $\begin{array}{c}\text{-8y - 8x + 16 = 0}\\ \text{8x = -8y + 16}\\ \text{x = -y + 2}\end{array}$ .

Substituting the value of $\text{x = -y + 2}$ in the first equation,

  $\begin{array}{c}{\text{(2 - y)}}^{\text{2}}{\text{ - y}}^{\text{2}}\text{ - 8(2 - y) + 8 = 0}\\ {\text{4 - 4y + y}}^{\text{2}}{\text{ - y}}^{\text{2}}\text{ - 16 + 8y + 8 = 0}\\ \text{4y = 4}\\ \text{y = 1}\end{array}$

Substituting the value of $\text{y = 1}$ in $\text{x = -y + 2}$ to get,

  $\begin{array}{c}\text{x = -y + 2}\\ \text{= -1 + 2}\\ \text{x = 1}\end{array}$

The ship’s location for the given hyperbolic equation is $\text{(1, 1)}$ .

b.

To find: To find the ship’s position.

The ship is at the point $\left(\text{-4}\sqrt{\text{5}}\text{, }\frac{\text{6}\sqrt{\text{5}}}{\text{5}}\right)$ .

Given information: The two hyperbolic equations are $\text{xy - 24 = 0}$ and ${\text{x}}^{\text{2}}{\text{ - 25y}}^{\text{2}}\text{ + 100 = 0}$ .

Calculation:

The hyperbolic equations are

  $\begin{array}{c}\text{xy - 24 = 0 }\mathrm{......}\text{(1)}\\ {\text{x}}^{\text{2}}{\text{ - 25y}}^{\text{2}}\text{ + 100 = 0 }\mathrm{......}\text{(2)}\end{array}$

Rearrange the first equation to get

  $\text{x = }\frac{\text{24}}{\text{y}}$

Substituting the value of $\text{x = }\frac{\text{24}}{\text{y}}$ in equation 2

  $\begin{array}{c}{\text{x}}^{\text{2}}{\text{ - 25y}}^{\text{2}}\text{ + 100 = 0}\\ {\left(\frac{\text{24}}{\text{y}}\right)}^{\text{2}}{\text{ - 25y}}^{\text{2}}\text{ + 100 = 0}\\ \frac{\text{576}}{{\text{y}}^{\text{2}}}{\text{ - 25y}}^{\text{2}}\text{ + 100 = 0}\end{array}$

Multiply the whole equation with ${\text{y}}^{\text{2}}$ to get,

  ${\text{576 - 25y}}^{\text{4}}{\text{ + 100y}}^{\text{2}}\text{ = 0}$

Substitute ${\text{t = y}}^2$ to get,

  ${\text{576 - 25t}}^2\text{ + 100t = 0}$

Solving the above equation to get, ${\text{t}}_{\text{1}}\text{ = }\frac{\text{36}}{\text{5}}{\text{ and t}}_{\text{2}}\text{ = -}\frac{\text{16}}{\text{5}}$

Substituting the value of ${\text{t}}_{\text{1}}\text{ = }\frac{\text{36}}{\text{5}}$

  $\begin{array}{c}{\text{y}}^{\text{2}}\text{ = t}\\ \text{y = }\sqrt{\text{t}}\\ \text{y = }\frac{\text{6}\sqrt{\text{5}}}{\text{5}}\end{array}$

To find the value of $\text{x}$ , substitute the value of $\text{y = }\frac{\text{6}\sqrt{\text{5}}}{\text{5}}$

  $\begin{array}{l}\text{x = }\frac{\text{24}}{\text{y}}\\ \text{ x = 4}\sqrt{5}\end{array}$

Since, the ship is at the west of the y-axis, let take $\text{x = -4}\sqrt{5}$

Therefore, the ship’s location for the given hyperbolic equation is $\left(\text{-4}\sqrt{\text{5}}\text{, }\frac{\text{6}\sqrt{\text{5}}}{\text{5}}\right)$ .