-x^2+4y^2-2x-16y+11=0?

-x² + 4y² - 2x - 16y + 11 = 0
There is an x² term and a y² term, and they have different signs, so the graph is a hyperbola.
The y² is positive, so the hyperbola is vertical.

General equation for a vertical hyperbola:
 (y-k)²/a² - (x-h)²/b² = 1
with
 center (h,k)
 vertices (h,k±a)
 foci (h,k±c), c² = a²+b²
 asymptotes y = ±(a/b)x + k∓(a/b)h

regroup terms
4y² - 16y - x² - 2x = -11

factor out the leading coefficients
4(y² - 4y) - (x² + 2x) = -11

complete the squares
4(y² - 4y + 2²) - (x² + 2x + 1²) = -11 + 4·2² - 1²
4(y-2)² - (x+1)² = 4

divide by the constant term, 4
(y-2)² - (x+1)²/4 = 1
(y-2)²/1² - (x+1)²/2² = 1

center (-1,2)
vertices (-1,2±1) = (-1,1) and (-1,3)

c² = 1² + 2² = 5
foci (-1,2±c) = (-1,2+√5) and (-1,2-√5)

asymptotes
y = -½x + 3/2
y = ½x + 5/2