Find the standard form of the equation of the hyperbola with the given characteristics.
foci: $( \pm 5,0 )$ asymptotes: $y = \pm 3 x$
A) $\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 9 } = 1$
B) $\frac { y ^ { 2 } } { 25 } - \frac { x ^ { 2 } } { 9 } = 1$
C) $\frac { y ^ { 2 } } { \frac { 5 } { 2 } } - \frac { x ^ { 2 } } { \frac { 45 } { 2 } } = 1$
D) $\frac { x ^ { 2 } } { \frac { 5 } { 2 } } - \frac { y ^ { 2 } } { \frac { 45 } { 2 } } = 1$
E) $\frac { x ^ { 2 } } { \frac { 45 } { 2 } } - \frac { y ^ { 2 } } { \frac { 5 } { 2 } } = 1$

Question

Quizplus · Accepted Answer

Since the foci are located on the x-axis, we know that the standard form of the hyperbola will be of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, where $a$ represents the distance from the center of the hyperbola to either vertex, and $b$ represents the distance from the center of the hyperbola to either co-vertex. Also, since the slopes of the asymptotes are $\pm 3$ and the center of the hyperbola is at the origin, we know that the asymptotes have equations $y = \pm \frac{b}{a}x$, where $\frac{b}{a} = 3$. Solving this equation for $b$ gives us $b = 3a$, and substituting this into the equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ gives us $\frac{x^2}{a^2} - \frac{y^2}{9a^2} = 1$. Lastly, we know that the distance between each focus and the center is 5, so $2a = 10$ and $a = 5$. Substituting this into our equation gives us $\frac{x^2}{5^2} - \frac{y^2}{9 \cdot 5^2} = 1$, which simplifies to $\boxed{	extbf{(D) } \frac{x^2}{\frac{5}{2}} - \frac{y^2}{\frac{45}{2}} = 1}$.

Find the Standard Form of the Equation of the Hyperbola (±5,0)( \pm 5,0 )(±5,0)

Find the Standard Form of the Equation of the Hyperbola $( \pm 5,0 )$