Chapter 9.5, Problem 34PPS

To determine

To Find :Equation of the horizontal hyperbola centered at the origin

Answer to Problem 34PPS

  $\frac{x^2}{1225}-\frac{y^2}{14400}=1$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Distance bweteen two station is : }250miles\\ \text{Distance of airplane is }70miles\text{ farther from station B from station A}\end{array}$

Formula Used:

  $\begin{array}{l}{c}^2=a^2+b^2\\ \text{Hyperbola Standard equation: }\frac{x^2}{a^2}-\frac{y^{{}^2}}{b^2}=1\end{array}$

Calculation:

  $\begin{array}{l}\text{Let the location of the two station be foci and the centre is at origin and distance between 2 foci is }2c=250\text{ miles}\\ \text{Standard equation is:}\\ \Rightarrow \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\ \Rightarrow 2c=250\\ \Rightarrow c=125\\ \text{The distance of 70 miles is constant difference}\\ \Rightarrow 2a=70\\ \Rightarrow a=35\\ \text{use }{c}^2=a^2+b^2\text{ to find }{b}^2\\ \Rightarrow b^2=c^2-a^2\\ \Rightarrow b^2={125}^2-{35}^2=14400\\ \text{the equation is :}\\ \frac{x^2}{{35}^2}-\frac{y^2}{14400}=1\\ \frac{x^2}{1225}-\frac{y^2}{14400}=1\end{array}$

Conclusion:

Thus the equation of the hyperbola is $\frac{x^2}{1225}-\frac{y^2}{14400}=1$ .