## Calculus (3rd Edition)

The answers: $\begin{array}{*{20}{c}} {}&{standard{\ }position}\\ {focal{\ }v.}&{\left( { \pm a,0} \right) = \left( { \pm 6,0} \right)}\\ {conjugate{\ }v.}&{\left( {0, \pm b} \right) = \left( {0, \pm 4\sqrt 3 } \right)}\\ {foci}&{\left( { \pm c,0} \right) = \left( { \pm \sqrt {84} ,0} \right)}\\ {center}&{\left( {0,0} \right)}\\ {asymptotes}&{y = \pm \frac{{2\sqrt 3 }}{3}x} \end{array}\begin{array}{*{20}{c}} {translated{\ }position}\\ {\left( {5,5} \right),\left( { - 7,5} \right)}\\ {\left( { - 1,5 \pm 4\sqrt 3 } \right)}\\ {\left( { \pm \sqrt {84} - 1,5} \right)}\\ {\left( { - 1,5} \right)}\\ {y = \pm \frac{{2\sqrt 3 }}{3}x + 5 \pm \frac{{2\sqrt 3 }}{3}} \end{array}$
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: $\begin{array}{*{20}{c}} {}&{standard{\ }position}\\ {focal{\ }v.}&{\left( { \pm a,0} \right) = \left( { \pm 6,0} \right)}\\ {conjugate{\ }v.}&{\left( {0, \pm b} \right) = \left( {0, \pm 4\sqrt 3 } \right)}\\ {foci}&{\left( { \pm c,0} \right) = \left( { \pm \sqrt {84} ,0} \right)}\\ {center}&{\left( {0,0} \right)}\\ {asymptotes}&{y = \pm \frac{{2\sqrt 3 }}{3}x} \end{array}\begin{array}{*{20}{c}} {translated{\ }position}\\ {\left( {5,5} \right),\left( { - 7,5} \right)}\\ {\left( { - 1,5 \pm 4\sqrt 3 } \right)}\\ {\left( { \pm \sqrt {84} - 1,5} \right)}\\ {\left( { - 1,5} \right)}\\ {y = \pm \frac{{2\sqrt 3 }}{3}x + 5 \pm \frac{{2\sqrt 3 }}{3}} \end{array}$