A nuclear cooling tower cross section is a hyperbola having a diameter of 52 ft at the center. The distance between the two foci is 96 ft. What is the equation for the hyperbola?
A) $\frac { x ^ { 2 } } { 676 } - \frac { y ^ { 2 } } { 1628 } = 5$
B) $\frac { x ^ { 2 } } { 676 } - \frac { y ^ { 2 } } { 1628 } = 1$
C) $\frac { x ^ { 2 } } { 676 } - \frac { y ^ { 2 } } { 2304 } = 1$
D) $\frac { x ^ { 2 } } { 676 } - \frac { y ^ { 2 } } { 1630 } = 1$

Question

A nuclear cooling tower cross section is a hyperbola having a diameter of 52 ft at the center. The distance between the two foci is 96 ft. What is the equation for the hyperbola? 
A) $\frac { x ^ { 2 } } { 676 } - \frac { y ^ { 2 } } { 1628 } = 5$
B) $\frac { x ^ { 2 } } { 676 } - \frac { y ^ { 2 } } { 1628 } = 1$
C) $\frac { x ^ { 2 } } { 676 } - \frac { y ^ { 2 } } { 2304 } = 1$
D) $\frac { x ^ { 2 } } { 676 } - \frac { y ^ { 2 } } { 1630 } = 1$

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 at the center in this case, which is 52 ft, so $a = 26$), and the distance between the foci is $2c = 96$, so $c = 48$. Using the relationship $c^2 = a^2 + b^2$, we find $b^2 = c^2 - a^2 = 48^2 - 26^2 = 2304 - 676 = 1628$. Therefore, the correct equation is $\frac{x^2}{676} - \frac{y^2}{1628} = 1$.

A Nuclear Cooling Tower Cross Section Is a Hyperbola Having x2676−y21628=5\frac { x ^ { 2 } } { 676 } - \frac { y ^ { 2 } } { 1628 } = 5676x2​−1628y2​=5

A Nuclear Cooling Tower Cross Section Is a Hyperbola Having $\frac { x ^ { 2 } } { 676 } - \frac { y ^ { 2 } } { 1628 } = 5$