Complete the square to write the equation [tex]\(4x^2 - 9y^2 + 24x + 18y - 9 = 0\)[/tex] in the standard form shown below, and then identify the key features of the graph.

[tex]\[
\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1
\][/tex]

- [tex]\(h = \square\)[/tex]
- [tex]\(k = \square\)[/tex]
- Center: [tex]\(\square\)[/tex]
- Slope of asymptote: [tex]\(\square\)[/tex]
- [tex]\(b = \square\)[/tex]



Answer :

To rewrite the equation [tex]\(4x^2 - 9y^2 + 24x + 18y - 9 = 0\)[/tex] in standard form and identify the key features of the graph, follow these steps:

### Step 1: Group and Complete the Square

Group the x and y terms:
[tex]\[ 4x^2 + 24x - 9y^2 + 18y - 9 = 0 \][/tex]

#### For [tex]\(x\)[/tex]-Terms

Isolate and factor out the coefficient of [tex]\(x^2\)[/tex]:
[tex]\[ 4(x^2 + 6x) \][/tex]

To complete the square for [tex]\( x^2 + 6x \)[/tex]:
1. Take half the coefficient of [tex]\( x \)[/tex]: [tex]\( \frac{6}{2} = 3 \)[/tex]
2. Square it: [tex]\( 3^2 = 9 \)[/tex]

So, we rewrite [tex]\( x^2 + 6x \)[/tex] as:
[tex]\[ x^2 + 6x + 9 - 9 \][/tex]

Now we factor and adjust:
[tex]\[ 4(x^2 + 6x + 9 - 9) = 4((x + 3)^2 - 9) = 4(x + 3)^2 - 36 \][/tex]

#### For [tex]\(y\)[/tex]-Terms

Isolate and factor out the coefficient of [tex]\( y^2 \)[/tex]:
[tex]\[ -9(y^2 - 2y) \][/tex]

To complete the square for [tex]\( y^2 - 2y \)[/tex]:
1. Take half the coefficient of [tex]\( y \)[/tex]: [tex]\( \frac{-2}{2} = -1 \)[/tex]
2. Square it: [tex]\((-1)^2 = 1 \)[/tex]

So, we rewrite [tex]\( y^2 - 2y \)[/tex] as:
[tex]\[ y^2 - 2y + 1 - 1 \][/tex]

Now we factor and adjust:
[tex]\[ -9(y^2 - 2y + 1 - 1) = -9((y - 1)^2 - 1) = -9(y - 1)^2 + 9 \][/tex]

### Step 2: Combine All Parts

Combine the completed squares:
[tex]\[ 4(x + 3)^2 - 36 - 9(y - 1)^2 + 9 - 9 = 0 \][/tex]
Simplify:
[tex]\[ 4(x + 3)^2 - 9(y - 1)^2 - 36 + 9 - 9 = 0 \][/tex]
[tex]\[ 4(x + 3)^2 - 9(y - 1)^2 - 36 = 0 \][/tex]

Move the constant to the other side and simplify:
[tex]\[ 4(x + 3)^2 - 9(y - 1)^2 = 36 \][/tex]

Divide both sides by 36:
[tex]\[ \frac{4(x + 3)^2}{36} - \frac{9(y - 1)^2}{36} = \frac{36}{36} \][/tex]
[tex]\[ \frac{(x + 3)^2}{9} - \frac{(y - 1)^2}{4} = 1 \][/tex]

Here, the equation is now in standard form.

### Step 3: Identify Key Features

Center (h, k):
In the standard form [tex]\(\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\)[/tex], we have:
[tex]\[ \frac{(x - (-3))^2}{9} - \frac{(y - 1)^2}{4} = 1 \][/tex]

So, the center is:
[tex]\[h = -3, \quad k = 1\][/tex]

Values of [tex]\(a^2\)[/tex] and [tex]\(b^2\)[/tex]:
[tex]\[a^2 = 9, \quad b^2 = 4\][/tex]

Slope of the Asymptotes:
The slopes of the asymptotes of the hyperbola given by [tex]\(\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\)[/tex] are given by [tex]\(\pm \frac{b}{a}\)[/tex]:
[tex]\[ \text{slope of asymptotes} = \pm \frac{b}{a} = \pm \frac{\sqrt{4}}{\sqrt{9}} = \pm \frac{2}{3} \][/tex]

### Summary of Key Features:

- Center:
[tex]\[h = -3\][/tex]
[tex]\[k = 1\][/tex]

- Values:
[tex]\[a^2 = 9\][/tex]
[tex]\[b^2 = 4\][/tex]

- Slope of the asymptotes:
[tex]\[ \text{slope} = \pm \frac{2}{3} \][/tex]

### Answer:
In conclusion, the key features are:
[tex]\[ h = -3, \quad k = 1, \][/tex]
[tex]\[ a^2 = 9, \quad b = \sqrt{4} = 2,\][/tex]
[tex]\[ \text{slope of asymptotes} = \pm \frac{2}{3} \][/tex]