Answer
Center: $(2,0)$
Foci: $(0,\pm \sqrt{10})$
Vertices: $(2, \pm 2\sqrt 2)$
Asymptote: $y=±2x \mp 4$
Work Step by Step
$y^2-4x^2+16x=24$
$y^2-4(x^2 -4x+ \textbf{4})=24 \textbf{ - 16}$
$y^2 -4(x-2)^2=8$
Hyperbola form: $\frac{y^2}{8} - \frac{4(x-2)^2}{8}=1$
$\qquad \qquad \qquad \enspace \frac{y^2}{8} - \frac{(x-2)^2}{2}=1$
$\qquad \qquad \qquad \enspace \frac{y^2}{(2\sqrt 2)^2} - \frac{(x-2)^2}{(\sqrt 2)^2}=1$
$\qquad \qquad \qquad \enspace \qquad \enspace h=2,k=0,a=\sqrt 2,b=2\sqrt 2$
Linear eccentricity (focal distance): $c= \sqrt{a^2+b^2}$
$\qquad \qquad \qquad \qquad \qquad \hspace{3em} c=\sqrt{(\sqrt2)^2+(2\sqrt2)^2}$
$\qquad \qquad \qquad \qquad \qquad \hspace{3em} c=\sqrt {10}$
Center: $(h,k)$
$\qquad \quad (2,0)$
Foci: $(h,k±c)$
$\qquad (2,0±\sqrt{10})$
$\qquad (2,\pm \sqrt{10})$
Vertices: $(h,k±b)$
$\qquad \quad \enspace (2, 0\pm 2\sqrt 2)$
$\qquad \quad \enspace (2, \pm 2\sqrt 2)$
Asymptote: $y=±\frac{b}{a}(x−h)+k$
$\qquad \qquad \enspace y=±\frac{2\sqrt 2}{\sqrt 2}(x-2)+0$
$\qquad \qquad \enspace y=±2x \mp 4$
