graph the hyperbola given by ( 9 x ^ { 2 } - y ^ { 2 } - 72 x - 4 y + 131 = 0 ).

Question

Accepted Answer

# Explanation:
## Step1: Complete the square for $x$ and $y$ terms
Group $x$ - terms and $y$ - terms:
$$
\begin{align*}
9x^{2}-72x-(y^{2}+4y)&=- 131\
9(x^{2}-8x)-(y^{2}+4y)&=-131
\end{align*}
$$
For the $x$ - part: $x^{2}-8x=(x - 4)^{2}-16$ (using $(a - b)^{2}=a^{2}-2ab + b^{2}$, here $a = x$, $b = 4$). For the $y$ - part: $y^{2}+4y=(y + 2)^{2}-4$ (using $(a + b)^{2}=a^{2}+2ab + b^{2}$, here $a=y$, $b = 2$)
$$
\begin{align*}
9((x - 4)^{2}-16)-((y + 2)^{2}-4)&=-131\
9(x - 4)^{2}-144-(y + 2)^{2}+4&=-131\
9(x - 4)^{2}-(y + 2)^{2}&=9
\end{align*}
$$
Divide by $9$ to get the standard form $\frac{(x - 4)^{2}}{1}-\frac{(y + 2)^{2}}{9}=1$

## Step2: Identify the center, $a$, $b$
The standard form of a hyperbola is $\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}}=1$, where $(h,k)$ is the center. Here $h = 4$, $k=-2$, $a = 1$, $b = 3$
The center of the hyperbola is $(4,-2)$, and it opens left - right (since the $x$ - term is positive)

# Answer:
The first graph (the one with the center near $(4,-2)$ and opening left - right) is the correct graph of the hyperbola $9x^{2}-y^{2}-72x - 4y+131 = 0$

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QUESTION IMAGE

Question

Explanation:

Step1: Complete the square for \(x\) and \(y\) terms

Step2: Identify the center, \(a\), \(b\)

Answer: