Chapter 9.5, Problem 31PPS

To determine

To Find :Equation of the path.

Answer to Problem 31PPS

  $\frac{x^2}{100}-\frac{y^2}{1250}=1$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Coordinates of the path are }\left(10,0\right)\text{ and }\left(30,100\right)\\ \text{Centre is at origin}\end{array}$

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:

  $\begin{array}{l}\text{To find }{a}^2\text{ and }{b}^2\text{ we use the standard equation of hyperbola: }\frac{x^2}{a^2}-\frac{y^{{}^2}}{b^2}=1\\ \text{Using coordinate }\left(10,0\right)\text{ to get }{a}^2\text{:}\\ \Rightarrow \frac{{10}^2}{a^2}-\frac{0^2}{b^2}=1\\ \Rightarrow \frac{100}{a^2}=1\\ \Rightarrow a^2=100\\ \text{Using coordinate }\left(30,100\right)\text{ to get }{b}^2\text{:}\\ \Rightarrow \frac{{30}^2}{100}-\frac{{100}^2}{b^2}=1\\ \Rightarrow 9-\frac{10000}{b^2}=1\\ \Rightarrow 8=\frac{10000}{b^2}\\ \Rightarrow b^2=1250\\ \text{Equation of hyperbola is: }\frac{x^2}{100}-\frac{y^2}{1250}=1\end{array}$

Conclusion:

Thus the equation of the path is $\frac{x^2}{100}-\frac{y^2}{1250}=1$