Decide if the equation defines an ellipse, a hyperbola, a parabola, or no conic section at all.displaystyle{left({a}right)}{4}{x}{2}-{9}{y}{2}={12}{left({b}right)}-{4}{x}+{9}{y}{2}={0}displaystyle{left({c}right)}{4}{y}{2}+{9}{x}{2}={12}{left({d}right)}{4}{x}{3}+{9}{y}{3}={12}

Standard equation of ellipse:
${\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1}$
Standard equation of a parabola:
${\displaystyle y^2=4ax}$
Standard equation of a Hyperbola:
${\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1}$
a)
${\displaystyle 4x^2-9y^2=12}$
Divide by coefficient of square terms : 4
${\displaystyle x^2-\frac{9}{4}{y}^2=3}$
Divide by coefficient of square terms : 9
${\displaystyle \frac{1}{9}{x}^2-\frac{1}{4}{y}^2=\frac{1}{3}}$
Divide by ${\displaystyle \frac{1}{3}}$
${\displaystyle \frac{x^2}{3}-\frac{y^2}{\frac{4}{3}}=1}$
So, this is the form of hyperbola ${\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1}$
Thus, the equation ${\displaystyle 4x^2-9y^2=12}$ defines a hyperbola.
b)
${\displaystyle -4x+9y^2=0}$
${\displaystyle 9y^2=4x}$
${\displaystyle y^2=4\frac{x}{9}}$
So, this is the form of parabola ${\displaystyle y^2=4ax}$
Thus, the equation ${\displaystyle -4x+9y^2=0}$ defines a parabola.
c)
${\displaystyle 4y^2+9x^2=12}$
Divide by coefficient of square terms : 9
${\displaystyle x^2+\frac{4}{9}{y}^2=\frac{4}{3}}$
Divide by coefficient of square terms : 4
${\displaystyle \frac{1}{4}{x}^2+\frac{1}{9}{y}^2=\frac{1}{3}}$
Divide by ${\displaystyle \frac{1}{3}}$
${\displaystyle \frac{x^2}{\frac{4}{3}}+\frac{y^2}{3}=1}$
This is the form of ellipse ${\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1}$
Thus, the equation ${\displaystyle 4y^2+9x^2=12}$ defines an ellipse.
d)
${\displaystyle 4x^3+9y^3=12}$
The above equation is not an ellipse, parabola and a hyperbola.
Hence, the equation ${\displaystyle 4x^3+9y^3=12}$ is not a conic section.