Answer
See below
Work Step by Step
Given $x^2-4y^2+8x-24y-24=0$
We can see that $a=1\\b=0\\c=-4$
We will find the discriminant of the given equation $=b^2-4ac\\=0^2-4(1)(-4)\\=16$
Since $16\gt 0$ and $a\ne c$, the conic is a hyperbola.
To graph the hyperbola, first complete the squar:
$x^2-4y^2+8x-24y-24=0\\x^2+8x-4y^2-24y=24\\(x^2+8x+16)-4(y^2+6y+9)=4\\(x+4)^2-4(y+3)^2=4\\\frac{(x+4)^2}{4}-\frac{(y+3)^2}{1}=1$
From the equation, you can see that the center is at $(-4,-3)$ and the foci is at $(-4 \pm \sqrt 5,-3)$
