write the equation of the ellipse ( 4x^{2}+9y^{2}+72x + 108y+612 = 0 ) in standard form.
( (x + 9)^{2}+(y + 6)^{2}=36 )
( \frac{(x + 9)^{2}}{9}+\frac{(y + 6)^{2}}{4}=1 )
( \frac{(x + 9)^{2}}{4}+\frac{(y + 6)^{2}}{9}=1 )
( \frac{(x - 9)^{2}}{9}+\frac{(y - 6)^{2}}{4}=1 )
question 19 (5 points)
write the equation for the hyperbola with foci ( (-12,6),(6,6) ) and vertices ( (-10,6),(4,6) ).
( \frac{(x + 3)^{2}}{7}-\frac{(y - 6)^{2}}{5.66}=1 )
( \frac{(x + 3)^{2}}{49}-\frac{(y - 6)^{2}}{32}=1 )
( \frac{(y + 3)^{2}}{49}-\frac{(x - 6)^{2}}{32}=1 )
( \frac{(x + 3)^{2}}{49}+\frac{(y - 6)^{2}}{32}=1 )

Question

Accepted Answer

# Explanation:
## Step1: Group $x$ and $y$ terms
$$
\begin{align*}
4x^{2}+9y^{2}+72x + 108y+612&=0\
4(x^{2}+18x)+9(y^{2}+12y)&=- 612
\end{align*}
$$
## Step2: Complete the square for $x$ and $y$
For $x$: $(x + 9)^{2}=x^{2}+18x + 81$, so $x^{2}+18x=(x + 9)^{2}-81$
For $y$: $(y + 6)^{2}=y^{2}+12y+36$, so $y^{2}+12y=(y + 6)^{2}-36$

Substitute back:
$$
\begin{align*}
4((x + 9)^{2}-81)+9((y + 6)^{2}-36)&=-612\
4(x + 9)^{2}-324+9(y + 6)^{2}-324&=-612\
4(x + 9)^{2}+9(y + 6)^{2}&=324 + 324-612\
4(x + 9)^{2}+9(y + 6)^{2}&=36
\end{align*}
$$
## Step3: Divide by 36 to get standard form
$$
\begin{align*}
\frac{4(x + 9)^{2}}{36}+\frac{9(y + 6)^{2}}{36}&=1\
\frac{(x + 9)^{2}}{9}+\frac{(y + 6)^{2}}{4}&=1
\end{align*}
$$

# Answer:
$\frac{(x + 9)^{2}}{9}+\frac{(y + 6)^{2}}{4}=1$ (the second option)

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QUESTION IMAGE

Question

Explanation:

Step1: Group \(x\) and \(y\) terms

Step2: Complete the square for \(x\) and \(y\)

Step3: Divide by 36 to get standard form

Answer: