# 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}

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

2021-02-23
Step 1
We need to determine the shape of the given conic section using its equation.
$$\displaystyle{x}^{2}-{2}{x}{y}+{y}^{2}+{x}-{2}{y}-{1}={0}$$
Step 2
We know that, for the general equation of conic section
$$\displaystyle{A}{x}^{2}+{B}{x}{y}+{C}{y}^{2}+{D}{x}+{E}{y}+{F}={0}$$
if $$\displaystyle{B}^{2}-{4}{A}{C}={0},$$ then the conic must be a parabola.
For the given equation, $$\displaystyle{A}={1},{B}=-{2},{C}={1}$$
$$\displaystyle{B}^{2}-{4}{A}{C}={\left(-{2}\right)}^{2}-{4}{\left({1}\right)}{\left({1}\right)}$$
$$\displaystyle={0}=>$$ given conic section is a parabola
Step 3
The graph of the parabola has been shown below.