Question

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

Answer 1

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.