Solve the problem.
-A nuclear cooling tower cross section is a hyperbola having a diameter of $54 \mathrm { ft }$ at the center. The distance between the two foci is $92 \mathrm { ft }$. What is the equation for the hyperbola?
A) $\frac { x ^ { 2 } } { 729 } - \frac { y ^ { 2 } } { 1387 } = 5$
B) $\frac { x ^ { 2 } } { 729 } - \frac { y ^ { 2 } } { 1389 } = 1$
C) $\frac { x ^ { 2 } } { 729 } - \frac { y ^ { 2 } } { 2116 } = 1$
D) $\frac { x ^ { 2 } } { 729 } - \frac { y ^ { 2 } } { 1387 } = 1$

Question

Quizplus · Accepted Answer

The distance between the foci is 2c, where c is the distance from the center to a focus. So, $2c = 92 \mathrm { ft }$, and $c = 46 \mathrm { ft }$. The diameter is 2a, where a is the distance from the center to a vertex. So, $2a = 54 \mathrm { ft }$, and $a = 27 \mathrm { ft }$. The equation for a hyperbola is $\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1$, where $c ^ { 2 } = a ^ { 2 } + b ^ { 2 }$. So, $b ^ { 2 } = c ^ { 2 } - a ^ { 2 } = 46 ^ { 2 } - 27 ^ { 2 } = 1387$. Therefore, the equation for the hyperbola is $\frac { x ^ { 2 } } { 729 } - \frac { y ^ { 2 } } { 1387 } = 1$.

Solve the Problem 54ft54 \mathrm { ft }54ft At the Center 92ft92 \mathrm { ft }92ft

Solve the Problem $54 \mathrm { ft }$ At the Center $92 \mathrm { ft }$