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

Tolnaio 2021-08-08 Answered
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
Answered 2021-08-09 Author has 22586 answers

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.

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