Answer :
To find the equations of the asymptotes for the given hyperbola [tex]\( 9x^2 - 4y^2 - 90x - 8y + 185 = 0 \)[/tex], we will proceed with the following steps:
### Step 1: Rearrange the Equation
First, group the [tex]\( x \)[/tex] and [tex]\( y \)[/tex] terms and move the constant term to the right side:
[tex]\[ 9x^2 - 4y^2 - 90x - 8y = -185 \][/tex]
### Step 2: Complete the Square
We'll complete the square for both [tex]\( x \)[/tex] and [tex]\( y \)[/tex] terms.
For the [tex]\( x \)[/tex]-terms:
[tex]\[ 9(x^2 - 10x) \][/tex]
[tex]\[ x^2 - 10x \][/tex]
To complete the square, add and subtract [tex]\( 25 \)[/tex] (since [tex]\( (10/2)^2 = 25 \)[/tex]):
[tex]\[ x^2 - 10x + 25 - 25 = (x - 5)^2 - 25 \][/tex]
So, we have:
[tex]\[ 9[(x - 5)^2 - 25] = 9(x - 5)^2 - 225 \][/tex]
For the [tex]\( y \)[/tex]-terms:
[tex]\[ -4(y^2 + 2y) \][/tex]
[tex]\[ y^2 + 2y \][/tex]
To complete the square, add and subtract [tex]\( 1 \)[/tex] (since [tex]\( (2/2)^2 = 1 \)[/tex]):
[tex]\[ y^2 + 2y + 1 - 1 = (y + 1)^2 - 1 \][/tex]
So, we have:
[tex]\[ -4[(y + 1)^2 - 1] = -4(y + 1)^2 + 4 \][/tex]
### Step 3: Substitute Back into the Equation
Substitute the squared forms back into the equation:
[tex]\[ 9[(x - 5)^2 - 25] - 4[(y + 1)^2 - 1] = -185 \][/tex]
[tex]\[ 9(x - 5)^2 - 225 - 4(y + 1)^2 + 4 = -185 \][/tex]
Combine the constants:
[tex]\[ 9(x - 5)^2 - 4(y + 1)^2 - 221 = -185 \][/tex]
Move the constant terms to the right side:
[tex]\[ 9(x - 5)^2 - 4(y + 1)^2 = 36 \][/tex]
### Step 4: Divide to Standard Form
Divide through by 36 to convert the equation into its standard form:
[tex]\[ \frac{9(x-5)^2}{36} - \frac{4(y+1)^2}{36} = \frac{36}{36} \][/tex]
[tex]\[ \frac{(x - 5)^2}{4} - \frac{(y + 1)^2}{9} = 1 \][/tex]
### Step 5: Identify Parameters
The standard form of a hyperbola with horizontal transverse axis is:
[tex]\[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \][/tex]
Compare and identify the parameters [tex]\( h \)[/tex], [tex]\( k \)[/tex], [tex]\( a^2 \)[/tex], and [tex]\( b^2 \)[/tex]:
[tex]\[ h = 5, \quad k = -1 \][/tex]
[tex]\[ a^2 = 4 \implies a = 2, \quad b^2 = 9 \implies b = 3 \][/tex]
### Step 6: Determine Asymptote Equations
The equations of the asymptotes for this standard form are:
[tex]\[ y - k = \pm\frac{b}{a}(x - h) \][/tex]
Substitute [tex]\( h = 5 \)[/tex], [tex]\( k = -1 \)[/tex], [tex]\( a = 2 \)[/tex], and [tex]\( b = 3 \)[/tex]:
[tex]\[ y + 1 = \pm\frac{3}{2}(x - 5) \][/tex]
Thus, we get two asymptote equations:
[tex]\[ y + 1 = \frac{3}{2}(x - 5) \][/tex]
[tex]\[ y + 1 = -\frac{3}{2}(x - 5) \][/tex]
So, the equations of the asymptotes for the given hyperbola are:
[tex]\[ y + 1 = \frac{3}{2}(x - 5) \][/tex]
[tex]\[ y + 1 = -\frac{3}{2}(x - 5) \][/tex]
### Step 1: Rearrange the Equation
First, group the [tex]\( x \)[/tex] and [tex]\( y \)[/tex] terms and move the constant term to the right side:
[tex]\[ 9x^2 - 4y^2 - 90x - 8y = -185 \][/tex]
### Step 2: Complete the Square
We'll complete the square for both [tex]\( x \)[/tex] and [tex]\( y \)[/tex] terms.
For the [tex]\( x \)[/tex]-terms:
[tex]\[ 9(x^2 - 10x) \][/tex]
[tex]\[ x^2 - 10x \][/tex]
To complete the square, add and subtract [tex]\( 25 \)[/tex] (since [tex]\( (10/2)^2 = 25 \)[/tex]):
[tex]\[ x^2 - 10x + 25 - 25 = (x - 5)^2 - 25 \][/tex]
So, we have:
[tex]\[ 9[(x - 5)^2 - 25] = 9(x - 5)^2 - 225 \][/tex]
For the [tex]\( y \)[/tex]-terms:
[tex]\[ -4(y^2 + 2y) \][/tex]
[tex]\[ y^2 + 2y \][/tex]
To complete the square, add and subtract [tex]\( 1 \)[/tex] (since [tex]\( (2/2)^2 = 1 \)[/tex]):
[tex]\[ y^2 + 2y + 1 - 1 = (y + 1)^2 - 1 \][/tex]
So, we have:
[tex]\[ -4[(y + 1)^2 - 1] = -4(y + 1)^2 + 4 \][/tex]
### Step 3: Substitute Back into the Equation
Substitute the squared forms back into the equation:
[tex]\[ 9[(x - 5)^2 - 25] - 4[(y + 1)^2 - 1] = -185 \][/tex]
[tex]\[ 9(x - 5)^2 - 225 - 4(y + 1)^2 + 4 = -185 \][/tex]
Combine the constants:
[tex]\[ 9(x - 5)^2 - 4(y + 1)^2 - 221 = -185 \][/tex]
Move the constant terms to the right side:
[tex]\[ 9(x - 5)^2 - 4(y + 1)^2 = 36 \][/tex]
### Step 4: Divide to Standard Form
Divide through by 36 to convert the equation into its standard form:
[tex]\[ \frac{9(x-5)^2}{36} - \frac{4(y+1)^2}{36} = \frac{36}{36} \][/tex]
[tex]\[ \frac{(x - 5)^2}{4} - \frac{(y + 1)^2}{9} = 1 \][/tex]
### Step 5: Identify Parameters
The standard form of a hyperbola with horizontal transverse axis is:
[tex]\[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \][/tex]
Compare and identify the parameters [tex]\( h \)[/tex], [tex]\( k \)[/tex], [tex]\( a^2 \)[/tex], and [tex]\( b^2 \)[/tex]:
[tex]\[ h = 5, \quad k = -1 \][/tex]
[tex]\[ a^2 = 4 \implies a = 2, \quad b^2 = 9 \implies b = 3 \][/tex]
### Step 6: Determine Asymptote Equations
The equations of the asymptotes for this standard form are:
[tex]\[ y - k = \pm\frac{b}{a}(x - h) \][/tex]
Substitute [tex]\( h = 5 \)[/tex], [tex]\( k = -1 \)[/tex], [tex]\( a = 2 \)[/tex], and [tex]\( b = 3 \)[/tex]:
[tex]\[ y + 1 = \pm\frac{3}{2}(x - 5) \][/tex]
Thus, we get two asymptote equations:
[tex]\[ y + 1 = \frac{3}{2}(x - 5) \][/tex]
[tex]\[ y + 1 = -\frac{3}{2}(x - 5) \][/tex]
So, the equations of the asymptotes for the given hyperbola are:
[tex]\[ y + 1 = \frac{3}{2}(x - 5) \][/tex]
[tex]\[ y + 1 = -\frac{3}{2}(x - 5) \][/tex]
