Identify the type of conic; then the equation to the standard form for that conic: x^2-2x-16y+33=0

y = (1/16)(x-1)² + 2

Step-by-step explanation:

There is exactly one squared variable, so the conic is a parabola.

The squared variable is x, not y, so the parabola is vertical.

Vertex-form equation for a vertical parabola:

y = a(x-h)²+k,

where (h,k) is the vertex, and a≠0.

Put x²-2x-16y+33=0 into vertex form:

y = (1/16)(x-1)² + 2

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