Question

# The equation is displaystyle{16}{x}^{2}+{9}{y}^{2}-{98}{x}+{5}{y}+{224}={0}

Conic sections
The equation is $$\displaystyle{16}{x}^{2}+{9}{y}^{2}-{98}{x}+{5}{y}+{224}={0}$$

2020-11-23

All conic section can be written in the general form
$$\displaystyle{A}{x}^{2}+{B}{x}{y}+{C}{y}^{2}+{D}{x}+{E}{y}+{F}={0}$$
The conic section represented by an equation in general form can be determine the coefficient.
If coefficient is $$\displaystyle{B}^{2}-{4}{A}{C}<{0},{B}={0}{\quad\text{and}\quad}{A}={C}$$, then its equation of circle.
If coefficient is $$\displaystyle{B}^{2}-{4}{A}{C}<{0},{B}\ne{0}{\quad\text{or}\quad}{A}\ne{C}$$, then its equation of ellipse
If coefficient is $$\displaystyle{B}^{2}-{4}{A}{C}>{0}$$, then its equation of hyperbola
If coefficicent is $$\displaystyle{B}^{2}-{4}{A}{C}={0}$$, then its equation of ellipse
To find value of $$\displaystyle{B}^{2}-{4}{A}{C}$$
Compare general equation with given equation
Here $$\displaystyle{A}={16},{B}={0},{C}={9},{D}=-{98},{E}={5}{\quad\text{and}\quad}{F}={224}$$
$$\displaystyle{B}^{2}-{4}{A}{C}={0}^{2}-{4}{\left({16}\right)}{\left({9}\right)}=-{576}$$
Here $$\displaystyle{B}^{2}-{4}{A}{C}<{0},{\quad\text{and}\quad}{A}\ne{C}$$
Therefore ,the given equation as an ellipse.