Use the eccentricity to find the equation of the hyperbola in standard form. eccentricity 2, foci (12, 0)and ( - 12, 0)

Question

Quizplus · Accepted Answer

The standard form of the hyperbola with center at the origin is:
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
The distance between the center and each focus is equal to $c$, where $c$ is the distance between each focus and the vertex.
In this case, the center is at $(0,0)$, the foci are at $(12,0)$ and $(-12,0)$, and the eccentricity is 2. 
Since $eccentricity = \frac{c}{a}$, we have $2 = \frac{c}{a}$. 
Also, we know that $c = 12$, so $a = \frac{c}{2} = 6$. 
Using $a$ and $b$, we can write the equation of the hyperbola in standard form:
$\frac{x^2}{36}-\frac{y^2}{b^2}=1$
We only need to find $b$. The distance between each vertex and each focus is given by $ae$, where $e$ is the eccentricity. In this case, $ae = 2a = 12$. Also, the distance between each vertex and each focus is equal to $b$. So, $b=12$. 
The final equation in standard form is:
$\frac{x^2}{36}-\frac{y^2}{144}=1$
Therefore, the best choice is (C).

Use the Eccentricity to Find the Equation of the Hyperbola