# The type of conic sections for the nondegenerate equations given below. a) 0.1x^{2}+0.6x-1.6=0.2y-0.1y^{2} b) 2x^{2}-7xy=-y^{2}+4x-2y-1

The type of conic sections for the nondegenerate equations given below.
a) $$\displaystyle{0.1}{x}^{{{2}}}+{0.6}{x}-{1.6}={0.2}{y}-{0.1}{y}^{{{2}}}$$
b) $$\displaystyle{2}{x}^{{{2}}}-{7}{x}{y}=-{y}^{{{2}}}+{4}{x}-{2}{y}-{1}$$
c) $$\displaystyle{8}{x}+{2}{y}={y}^{{{2}}}+{4}$$

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Delorenzoz

a) Consider the equation $$\displaystyle{0.1}{x}^{{{2}}}+{0.6}{x}-{1.6}={0.2}{y}-{0.1}{y}^{{{2}}}$$
This equation can be written as: $$0.1x^{2}+0.1y^{2}+0.6x-0.2y-1.6=0.$$
It is known that the equation is in the form $$\displaystyle{A}{x}^{{{2}}}+{C}{y}^{{{2}}}+{D}{x}+{E}{y}+{F}={0}.$$
Here, the coefficient of $$\displaystyle{x}^{{{2}}}$$ and $$\displaystyle{y}^{{{2}}}$$ are $$\displaystyle{A}={0.1}$$ and $$\displaystyle{C}={0.1}$$, that is, $$\displaystyle{A}={C}$$ Therefore, the equation is of a circle.
Hence, the conic section of the given equation is a circle.
b) Consider the equation $$\displaystyle{2}{x}^{{{2}}}-{7}{x}{y}=-{y}^{{{2}}}+{4}{x}-{2}{y}-{1}.$$
This equation can be written as: $$\displaystyle{2}{x}^{{{2}}}-{7}{x}{y}+{y}^{{{2}}}-{4}{x}+{2}{y}+{1}={0}$$. This is in the form of $$\displaystyle{A}{x}^{{{2}}}+{B}{x}{y}+{C}{y}^{{{2}}}+{D}{x}+{E}{y}+{F}={0}$$
Solve for $$\displaystyle{B}^{{{2}}}-{4}{A}{C}.$$
$$\displaystyle{B}^{{{2}}}-{4}{A}{C}={\left(\overline{{{7}}}-{4}{\left({2}\right)}{\left({1}\right)}\right)}$$
$$\displaystyle={49}-{8}$$
$$\displaystyle={41}$$
This implies that $$\displaystyle{B}^{{{2}}}-{4}{A}{C}{>}{0}$$ Therefore, the equation is of a hyperbola.
Hence, the conic section of the given equation is a hyperbola.
c) Consider the equation $$\displaystyle{8}{x}+{2}{y}={y}^{{{2}}}+{4}.$$
This equation can be written as: $$\displaystyle-{y}^{{{2}}}+{8}{x}+{2}{y}-{4}={0}$$
It is known that the equation is in the form $$\displaystyle{A}{x}^{{{2}}}+{C}{y}+{D}{x}+{E}{y}+{F}={0}$$
In the given equation, there is no term of $$\displaystyle{x}^{{{2}}},$$ which means that $$\displaystyle{A}={0}$$, that is, $$\displaystyle{A}{C}={0}$$ Therefore, the equation is of a parabola.
Hence, the conic section of the given equation is a parabola.