Chapter 9 Quadratic Relations and Conic Sections - 9.5 Graph and Write Equations of Hyperbolas - 9.5 Exercises - Problem Solving - Page 647: 41b

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We know A is $30.5$ ft from the origin, so $a=30.5$ The standard form for a horizontal transverse hyperbola is: $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\\frac{x^2}{(30.5)^2}-\frac{y^2}{b^2}=1\\\frac{x^2}{930.25}-\frac{y^2}{b^2}=1$$ Substitute $B(85,-40)$ to solve for b: $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\\frac{85^2}{930.25}-\frac{(-40)^2}{b^2}=1\\b^2=\frac{(-40)^2}{\frac{85^2}{930.25}-1}\approx 236.451$$ Hence, $$\frac{x^2}{930.25}-\frac{y^2}{236.451}=1$$