Answer
See below
Work Step by Step
Given $y^2+14y+16x+33=0$
We can see that $a=0\\b=0\\c=1$
We will find the discriminant of the given equation $=b^2-4ac\\=0^2-4(0)(1)\\=0$
The conic is a hyperbola.
To graph the hyperbola, first complete the square:
$y^2+14y+16x+33=0\\y^2+14y=-16x-33\\(y^2+14y+49)-49=-16x-33\\(y+7)^2=-16x+16\\(y+7)^2=-16(x-1)$
From the equation, you can see that the vertex is at $(1,-7)$
