A nuclear cooling tower cross section is a hyperbola having a diameter of 56 ft at the center. The distance between the two foci is 100 ft. What is the equation for the hyperbola?
A) $\frac { x ^ { 2 } } { 784 } - \frac { y ^ { 2 } } { 1716 } = 1$
B) $\frac { x ^ { 2 } } { 784 } - \frac { y ^ { 2 } } { 2500 } = 1$
C) $\frac { x ^ { 2 } } { 784 } - \frac { y ^ { 2 } } { 1718 } = 1$
D) $\frac { x ^ { 2 } } { 784 } - \frac { y ^ { 2 } } { 1716 } = 5$

Question

A nuclear cooling tower cross section is a hyperbola having a diameter of 56 ft at the center. The distance between the two foci is 100 ft. What is the equation for the hyperbola? 
A) $\frac { x ^ { 2 } } { 784 } - \frac { y ^ { 2 } } { 1716 } = 1$
B) $\frac { x ^ { 2 } } { 784 } - \frac { y ^ { 2 } } { 2500 } = 1$
C) $\frac { x ^ { 2 } } { 784 } - \frac { y ^ { 2 } } { 1718 } = 1$
D) $\frac { x ^ { 2 } } { 784 } - \frac { y ^ { 2 } } { 1716 } = 5$

Quizplus · Accepted Answer

The equation of a hyperbola in standard form is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, where $2a$ is the length of the transverse axis (diameter in this context) and $2b$ is the length of the conjugate axis. The distance between the foci is $2c$, where $c^2 = a^2 + b^2$. Given $2a = 56$, we have $a = 28$ and $a^2 = 784$. Given the distance between the foci $2c = 100$, we have $c = 50$ and $c^2 = 2500$. Using $c^2 = a^2 + b^2$, we find $b^2 = c^2 - a^2 = 2500 - 784 = 1716$. Therefore, the correct equation is $\frac{x^2}{784} - \frac{y^2}{1716} = 1$.

A Nuclear Cooling Tower Cross Section Is a Hyperbola Having x2784−y21716=1\frac { x ^ { 2 } } { 784 } - \frac { y ^ { 2 } } { 1716 } = 1784x2​−1716y2​=1

A Nuclear Cooling Tower Cross Section Is a Hyperbola Having $\frac { x ^ { 2 } } { 784 } - \frac { y ^ { 2 } } { 1716 } = 1$