Question

# The type of conic sections for the nondegenerate equations given below. a) 6x^{2} + 3x + 10y=10y^{2} + 8 b) 3x^{2} + 18xy=5x + 2y + 9 c) 4x^{2} + 8x - 5= -y^{2} + 6y + 3

Conic sections
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) $$\displaystyle{4}{x}^{{{2}}}\ +\ {8}{x}\ -\ {5}=\ -{y}^{{{2}}}\ +\ {6}{y}\ +\ {3}$$

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, $$\displaystyle{A}{C}\ {<}\ {0}$$</span>. 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 $$\displaystyle{4}{x}^{{{2}}}\ +\ {8}{x}\ -\ {5}=\ -{y}^{{{2}}}\ +\ {6}{y}\ +\ {3}$$. This 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.