Find the standard-form equation for a hyperbola which satisfies the given conditions.
-A hyperbola centered at the origin having vertex at (0, -8) and eccentricity equal to 2
A) $\frac { y ^ { 2 } } { 64 } - \frac { x ^ { 2 } } { 192 } = 1$
B) $\frac { y ^ { 2 } } { 64 } - \frac { x ^ { 2 } } { 64 } = 1$
C) $\frac { y ^ { 2 } } { 64 } + \frac { x ^ { 2 } } { 192 } = 1$
D) $\frac { x ^ { 2 } } { 64 } - \frac { y ^ { 2 } } { 192 } = 1$

Question

Find the standard-form equation for a hyperbola which satisfies the given conditions.
-A hyperbola centered at the origin having vertex at (0, -8) and eccentricity equal to 2 
A) $\frac { y ^ { 2 } } { 64 } - \frac { x ^ { 2 } } { 192 } = 1$
B) $\frac { y ^ { 2 } } { 64 } - \frac { x ^ { 2 } } { 64 } = 1$
C) $\frac { y ^ { 2 } } { 64 } + \frac { x ^ { 2 } } { 192 } = 1$
D) $\frac { x ^ { 2 } } { 64 } - \frac { y ^ { 2 } } { 192 } = 1$

Quizplus · Accepted Answer

The standard form of a hyperbola centered at the origin is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ for a horizontal hyperbola or $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ for a vertical hyperbola. Given the vertex is at (0, -8), this indicates a vertical hyperbola, so $a = 8$ and $a^2 = 64$. The eccentricity $e = \frac{c}{a}$, where $c$ is the distance from the center to a focus. Given $e = 2$, we have $2 = \frac{c}{8}$, so $c = 16$. Using $c^2 = a^2 + b^2$, we find $b^2 = c^2 - a^2 = 256 - 64 = 192$. Thus, the equation is $\frac{y^2}{64} - \frac{x^2}{192} = 1$.

Find the Standard-Form Equation for a Hyperbola Which Satisfies the Given