 # The equation is displaystyle{16}{x}^{2}+{9}{y}^{2}-{98}{x}+{5}{y}+{224}={0} banganX 2020-11-22 Answered
The equation is $16{x}^{2}+9{y}^{2}-98x+5y+224=0$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it hesgidiauE

All conic section can be written in the general form
$A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$
The conic section represented by an equation in general form can be determine the coefficient.
If coefficient is ${B}^{2}-4AC<0,B=0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}A=C$, then its equation of circle.
If coefficient is ${B}^{2}-4AC<0,B\ne 0\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}A\ne C$, then its equation of ellipse
If coefficient is ${B}^{2}-4AC>0$, then its equation of hyperbola
If coefficicent is ${B}^{2}-4AC=0$, then its equation of ellipse
To find value of ${B}^{2}-4AC$
Compare general equation with given equation
Here $A=16,B=0,C=9,D=-98,E=5\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}F=224$
${B}^{2}-4AC={0}^{2}-4\left(16\right)\left(9\right)=-576$
Here ${B}^{2}-4AC<0,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}A\ne C$
Therefore ,the given equation as an ellipse.