Answer
See below
Work Step by Step
Given: $9x^2-y^2-18x-4y-5=0$
Because $A=9,B=0$, and $C=1$, the discriminant is $B^2-4AC=-96 \lt 0$, so the conic is a hyperbola. Complete the square to write the equation in standard form.
$9x^2-y^2-18x-4y-5=0\\9(x^2-2x)-(y^2+4y)=5\\9(x^2-2x+1-1)-(y^2+4y+4-4)=5\\9(x^2-2x+1)-(y^2+4y+4)=10\\9(x-1)^2-(y+2)^2=10\\\frac{9(x-1)^2}{10}-\frac{(y+2)^2}{10}=1$
From the equation, we can see the center is at $(1,-2)$. The graph is shown below.
