Answer
The answers:
$\begin{array}{*{20}{c}}
{}&{standard{\ }position}\\
{focal{\ }v.}&{\left( { \pm a,0} \right) = \left( { \pm 6,0} \right)}\\
{conjugate{\ }v.}&{\left( {0, \pm b} \right) = \left( {0, \pm 4\sqrt 3 } \right)}\\
{foci}&{\left( { \pm c,0} \right) = \left( { \pm \sqrt {84} ,0} \right)}\\
{center}&{\left( {0,0} \right)}\\
{asymptotes}&{y = \pm \frac{{2\sqrt 3 }}{3}x}
\end{array}\begin{array}{*{20}{c}}
{translated{\ }position}\\
{\left( {5,5} \right),\left( { - 7,5} \right)}\\
{\left( { - 1,5 \pm 4\sqrt 3 } \right)}\\
{\left( { \pm \sqrt {84} - 1,5} \right)}\\
{\left( { - 1,5} \right)}\\
{y = \pm \frac{{2\sqrt 3 }}{3}x + 5 \pm \frac{{2\sqrt 3 }}{3}}
\end{array}$
Work Step by Step
Write
$4{x^2} - 3{y^2} + 8x + 30y = 215$
$4{x^2} + 8x - \left( {3{y^2} - 30y} \right) = 215$
$4\left( {{x^2} + 2x} \right) - 3\left( {{y^2} - 10y} \right) = 215$
$4\left( {{x^2} + 2x + 1} \right) - 4 - 3\left( {{y^2} - 10y + 25} \right) + 75 = 215$
So,
$4{\left( {x + 1} \right)^2} - 3{\left( {y - 5} \right)^2} = 144$
Divide both sides by 144 gives
$\frac{4}{{144}}{\left( {x + 1} \right)^2} - \frac{3}{{144}}{\left( {y - 5} \right)^2} = 1$
${\left( {\frac{{x + 1}}{6}} \right)^2} - {\left( {\frac{{y - 5}}{{12/\sqrt 3 }}} \right)^2} = 1$
Or
${\left( {\frac{{x + 1}}{6}} \right)^2} - {\left( {\frac{{y - 5}}{{4\sqrt 3 }}} \right)^2} = 1$
From this equation we see that this is a hyperbola with $a=6$, $b = 4\sqrt 3 $ and centered at $C=\left(-1,5\right)$.
By Theorem 2, we get
$c = \sqrt {{a^2} + {b^2}} = \sqrt {36 + 48} = \sqrt {84} $
Translated vertices:
Focal vertices:
$\left(6,0\right)+\left(-1,5\right)=\left(5,5\right)$
$\left(-6,0\right)+\left(-1,5\right)=\left(-7,5\right)$
Conjugate vertices:
$\left( {0,4\sqrt 3 } \right) + \left( { - 1,5} \right) = \left( { - 1,5 + 4\sqrt 3 } \right)$
$\left( {0, - 4\sqrt 3 } \right) + \left( { - 1,5} \right) = \left( { - 1,5 - 4\sqrt 3 } \right)$
Translated foci:
$\left( {\sqrt {84} ,0} \right) + \left( { - 1,5} \right) = \left( {\sqrt {84} - 1,5} \right)$
$\left( { - \sqrt {84} ,0} \right) + \left( { - 1,5} \right) = \left( { - \sqrt {84} - 1,5} \right)$
Translated asymptotes:
Since the center is translated to $C=\left(-1,5\right)$, the asymptotes are shifted $1$ unit to the left and $5$ units up. Thus,
$y = \frac{{2\sqrt 3 }}{3}\left( {x + 1} \right) + 5 = \frac{{2\sqrt 3 }}{3}x + 5 + \frac{{2\sqrt 3 }}{3}$
$y = - \frac{{2\sqrt 3 }}{3}\left( {x + 1} \right) + 5 = - \frac{{2\sqrt 3 }}{3}x + 5 - \frac{{2\sqrt 3 }}{3}$
We summarize the results in the following table:
$\begin{array}{*{20}{c}}
{}&{standard{\ }position}\\
{focal{\ }v.}&{\left( { \pm a,0} \right) = \left( { \pm 6,0} \right)}\\
{conjugate{\ }v.}&{\left( {0, \pm b} \right) = \left( {0, \pm 4\sqrt 3 } \right)}\\
{foci}&{\left( { \pm c,0} \right) = \left( { \pm \sqrt {84} ,0} \right)}\\
{center}&{\left( {0,0} \right)}\\
{asymptotes}&{y = \pm \frac{{2\sqrt 3 }}{3}x}
\end{array}\begin{array}{*{20}{c}}
{translated{\ }position}\\
{\left( {5,5} \right),\left( { - 7,5} \right)}\\
{\left( { - 1,5 \pm 4\sqrt 3 } \right)}\\
{\left( { \pm \sqrt {84} - 1,5} \right)}\\
{\left( { - 1,5} \right)}\\
{y = \pm \frac{{2\sqrt 3 }}{3}x + 5 \pm \frac{{2\sqrt 3 }}{3}}
\end{array}$
