Sure, let's solve the problem step-by-step.
### Part (a): Write the standard equation of the hyperbola
Given the general form of the hyperbola:
[tex]\[ 4x^2 + 8x - 9y^2 - 36y - 68 = 0 \][/tex]
We need to convert this into standard form. The standard form for a hyperbola is:
[tex]\[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \][/tex]
or
[tex]\[ - \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \][/tex]
First, we complete the square for the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms. We'll separate the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms and move the constant term to the other side:
Step 1: Complete the square for the [tex]\(x\)[/tex] terms:
[tex]\[ 4x^2 + 8x \][/tex]
Factor out the 4:
[tex]\[ 4(x^2 + 2x) \][/tex]
Complete the square inside the parentheses:
[tex]\[ 4(x^2 + 2x + 1 - 1) = 4((x + 1)^2 - 1) = 4(x + 1)^2 - 4 \][/tex]
Step 2: Complete the square for the [tex]\(y\)[/tex] terms:
[tex]\[ -9y^2 - 36y \][/tex]
Factor out the -9:
[tex]\[ -9(y^2 + 4y) \][/tex]
Complete the square inside the parentheses:
[tex]\[ -9(y^2 + 4y + 4 - 4) = -9((y + 2)^2 - 4) = -9(y + 2)^2 + 36 \][/tex]
Step 3: Rewrite the equation:
Substitute these completed squares back into the original equation:
[tex]\[ 4(x + 1)^2 - 4 - 9(y + 2)^2 + 36 - 68 = 0 \][/tex]
Simplify:
[tex]\[ 4(x + 1)^2 - 9(y + 2)^2 - 36 = 0 \][/tex]
Step 4: Isolate the hyperbolic equation:
Move the constant to the other side:
[tex]\[ 4(x + 1)^2 - 9(y + 2)^2 = 36 \][/tex]
Step 5: Divide by 36 to get standard form:
[tex]\[ \frac{4(x + 1)^2}{36} - \frac{9(y + 2)^2}{36} = 1 \][/tex]
Simplify the fractions:
[tex]\[ \frac{(x + 1)^2}{9} - \frac{(y + 2)^2}{4} = 1 \][/tex]
So the standard form of the hyperbola is:
[tex]\[ \frac{(x + 1)^2}{9} - \frac{(y + 2)^2}{4} = 1 \][/tex]
### Part (b): What is the center of the hyperbola?
In the standard form of the hyperbola:
[tex]\[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \][/tex]
The center is [tex]\((h, k)\)[/tex].
Comparing this with our equation:
[tex]\[ \frac{(x + 1)^2}{9} - \frac{(y + 2)^2}{4} = 1 \][/tex]
We can see that [tex]\( x + 1 \)[/tex] corresponds to [tex]\( (x - (-1)) \)[/tex] and [tex]\( y + 2 \)[/tex] corresponds to [tex]\( (y - (-2)) \)[/tex].
Thus, the center of the hyperbola is:
[tex]\[ (-1, -2) \][/tex]
