To find out which hyperbola's asymptote rectangle has the greatest perimeter, we need to calculate the perimeter of the asymptote rectangle for each hyperbola. The formula for the perimeter [tex]\( P \)[/tex] of the asymptote rectangle is given by:
[tex]\[ P = 4(a + b) \][/tex]
where [tex]\( a \)[/tex] is the semi-major axis and [tex]\( b \)[/tex] is the semi-minor axis of the hyperbola.
Let's identify the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] for each hyperbola and compute their perimeters:
1. Hyperbola A: [tex]\(\frac{(z-4)^2}{11^2} - \frac{(y+2)^2}{3^2} = 1\)[/tex]
- [tex]\( a = 11 \)[/tex]
- [tex]\( b = 3 \)[/tex]
- Perimeter [tex]\( P = 4(a + b) = 4(11 + 3) = 4 \times 14 = 56 \)[/tex]
2. Hyperbola B: [tex]\(\frac{(x-2)^2}{4^2} - \frac{(y+1)^2}{10^2} = 1\)[/tex]
- [tex]\( a = 4 \)[/tex]
- [tex]\( b = 10 \)[/tex]
- Perimeter [tex]\( P = 4(a + b) = 4(4 + 10) = 4 \times 14 = 56 \)[/tex]
3. Hyperbola C: [tex]\(\frac{(x+5)^2}{5^2} - \frac{(y-3)^2}{9^2} = 1\)[/tex]
- [tex]\( a = 5 \)[/tex]
- [tex]\( b = 9 \)[/tex]
- Perimeter [tex]\( P = 4(a + b) = 4(5 + 9) = 4 \times 14 = 56 \)[/tex]
4. Hyperbola D: [tex]\(\frac{(x-7)^2}{8^2} - \frac{(y-2)^2}{7^2} = 1\)[/tex]
- [tex]\( a = 8 \)[/tex]
- [tex]\( b = 7 \)[/tex]
- Perimeter [tex]\( P = 4(a + b) = 4(8 + 7) = 4 \times 15 = 60 \)[/tex]
After calculating the perimeters, we can compare the values:
- Hyperbola A: 56
- Hyperbola B: 56
- Hyperbola C: 56
- Hyperbola D: 60
The hyperbola with the greatest perimeter of the asymptote rectangle is Hyperbola D with a perimeter of 60.
Therefore, the correct answer is:
D. [tex]\(\frac{(x-7)^2}{8^2} - \frac{(y-2)^2}{7^2} = 1\)[/tex]
