# Question #fdc59

Write the equation in standard form as ${\left(x - \frac{1}{3}\right)}^{2} / {1}^{2} - {\left(y - 3\right)}^{2} / {6}^{2}$=1. The transverse axis of this hyperbola is parallel to x axis and its center is (1/3,3). a=1 , b= 6
Since ${c}^{2} = {a}^{2} + {b}^{2}$, c= $\sqrt{37}$
The vertices are 'a' units from the center, hence one vertex would be $\left(\frac{4}{3} , 3\right)$ and the other would be $\left(- \frac{2}{3} , 3\right)$.
$\sqrt{37} + \frac{1}{3}$,3 and the other would be -$\sqrt{37} + \frac{1}{3}$,3
Asymptotes would be y= $3 + 6 \left(x - \frac{1}{3}\right) \mathmr{and} y = 3 - 6 \left(x - \frac{1}{3}\right)$, that is y= 6x+1 and y= -6x+5