In this equation $$2x²+y²-4x-6y+11=0$$ I got the result $(1,3)$ completing squares $2(x - 1)² + (y - 3)² = 0$
But on my list exercises, demanded that determine the foci, straight guideline asymptote, vertex, center, radius, minor axis and major axis (if there were any of them according to the data items) of the following equations of conics. So, how can I get the result of a conical? Thank you.
You are correct that your given equation defines precisely one point in the reals. And as DonAntonio suggests, a point can be considered a degenerate conic.
However, it is possible that there is a typo in your exercises, and/or in you post.
The equation, with one sign change, defines an hyperbola centered at $(1, -3)$:
Specifically, if we subtract the term $y^2$ we have
$$2x^2 - y^2 - 4x - 6y + 11 = 0\iff 2(x-1)^2 - (y+3)^2 + 18 = 0$$
Double check the assigned problem to see if you may have transcribed the problem incorrectly. Otherwise, if you have copied it correctly, I suspect there's a typo in the source.