Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin.
Vertices: $( \pm 2,0 )$ ;asymptotes: $y = \pm 7 x$
A) $\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 196 } = 1$
B) $\frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 196 } = 1$
C) $\frac { y ^ { 2 } } { 4 } + \frac { x ^ { 2 } } { 196 } = 1$
D) $\frac { y ^ { 2 } } { 196 } - \frac { x ^ { 2 } } { 4 } = 1$
E) $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 196 } = 1$

Question

Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin. ​
Vertices: $( \pm 2,0 )$ ;asymptotes: $y = \pm 7 x$ ​
A) $\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 196 } = 1$
B) $\frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 196 } = 1$
C) $\frac { y ^ { 2 } } { 4 } + \frac { x ^ { 2 } } { 196 } = 1$
D) $\frac { y ^ { 2 } } { 196 } - \frac { x ^ { 2 } } { 4 } = 1$
E) $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 196 } = 1$

Quizplus · Accepted Answer

The vertices at $(\pm 2,0)$ indicate that the major axis is along the x-axis, so the x-term will be positive in the standard form of the hyperbola equation. The distance from the center to a vertex is $a=2$, so $a^2=4$. The slope of the asymptotes for a hyperbola centered at the origin with a horizontal transverse axis is $\pm \frac{b}{a}$. Given the asymptotes are $y = \pm 7x$, we have $\frac{b}{a} = 7$. Since $a=2$, solving for $b$ gives $b=14$, so $b^2=196$. Therefore, the standard form is $\frac{x^2}{4} - \frac{y^2}{196} = 1$.

Find the Standard Form of the Equation of the Hyperbola (±2,0)( \pm 2,0 )(±2,0)

Find the Standard Form of the Equation of the Hyperbola $( \pm 2,0 )$