 # 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} midtlinjeg 2020-11-24 Answered

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

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