Given the following conic section, determine its shape and then sketch its graph. displaystyle{x}^{2}-{2}{x}{y}+{y}^{2}+{x}-{2}{y}-{1}={0}

Step 1
We need to determine the shape of the given conic section using its equation.
${\displaystyle x^2-2xy+y^2+x-2y-1=0}$
Step 2
We know that, for the general equation of conic section
${\displaystyle A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0}$
if ${\displaystyle B^2-4AC=0,}$ then the conic must be a parabola.
For the given equation, ${\displaystyle A=1,B=-2,C=1}$
${\displaystyle B^2-4AC={(-2)}^2-4\left(1\right)\left(1\right)}$
${\displaystyle =0=>}$ given conic section is a parabola
Step 3
The graph of the parabola has been shown below.