Question

The type of conic sections for the nondegenerate equations given below.

a) \(\displaystyle{6}{x}^{{{2}}}\ +\ {3}{x}\ +\ {10}{y}={10}{y}^{{{2}}}\ +\ {8}\)

b) \(\displaystyle{3}{x}^{{{2}}}\ +\ {18}{x}{y}={5}{x}\ +\ {2}{y}\ +\ {9}\)

c) \(\displaystyle{4}{x}^{{{2}}}\ +\ {8}{x}\ -\ {5}=\ -{y}^{{{2}}}\ +\ {6}{y}\ +\ {3}\)

Answer 1

a) Consider the equation \(\displaystyle{6}{x}^{{{2}}}\ +\ {3}{x}\ +\ {10}{y}={10}{y}^{{{2}}}\ +\ {8}.\)

This equation can be written as: \(\displaystyle{6}{x}^{{{2}}}\ -\ {10}{y}^{{{2}}}\ +\ {10}{y}\ -\ {8}={0}\).

It is known that the equation is in the form of \(\displaystyle{A}{x}^{{{2}}}\ +\ {C}{y}^{{{2}}}\ +\ {D}{x}\ +\ {E}{y}\ +\ {F}={0}\)

Here, the coefficient of \(\displaystyle{x}^{{{2}}}\) and \(\displaystyle{y}^{{{2}}}\) are \(\displaystyle{A}={6}\) and \(\displaystyle{C}=\ -{10}\), which means \(\displaystyle{A}{C}=\ -{60},\) that is, \(\displaystyle{A}{C}\ {<}\ {0}\).

Therefore, the equation is of a hyperbola. Hence, the conic section of the given equation is a hyperbola.

b) Consider the equation \(\displaystyle{3}{x}^{{{2}}}\ +\ {18}{x}{y}={5}{x}\ +\ {2}{y}\ +\ {9}\).

This equation caan be written as: \(\displaystyle{3}{x}^{{{2}}}\ +\ {18}{x}{y}\ -\ {5}{x}\ -\ {2}{y}\ -\ {9}={0}\), which is in the form of \(\displaystyle{A}{X}^{{{2}}}\ +\ {B}{x}{y}\ +\ {C}{y}^{{{2}}}\ +\ {D}{x}\ +\ {E}{y}\ +\ {F}={0}\).

Solve for \(\displaystyle{B}^{{{2}}}\ -\ {4}{A}{C}\)
\(\displaystyle{B}^{{{2}}}\ -\ {4}{A}{C}={\left({18}\right)}\ -\ {4}{\left({3}\right)}{\left({0}\right)}\)
\(\displaystyle={324}\ -\ {0}\)
\(\displaystyle={324}\)

This shows 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 heperbola.

c) Consider the equation \(\displaystyle{4}{x}^{{{2}}}\ +\ {8}{x}\ -\ {5}=\ -{y}^{{{2}}}\ +\ {6}{y}\ +\ {3}\). T

his equation can written as: \(\displaystyle{4}{x}^{{{2}}}\ +\ {y}^{{{2}}}\ +\ {8}{x}\ -\ {6}{y}\ -\ {8}={0}\). It is known that the equation is in the form \(\displaystyle{A}{X}^{{{2}}}\ +\ {C}{y}^{{{2}}}\ +\ {D}{x}\ +\ {E}{y}\ +\ {F}={0}\) Here, \(\displaystyle{A}={4}\ \text{and}\ {C}={1},\) which means \(\displaystyle{A}{C}={4}\ \text{and}\ {A}{C}\ {>}\ {0}.\)

Therefore, the equation is of an ellipse. Hence, the conic section of the given equation is an ellipse.