To determine which hyperbola has its foci farthest from its center, we need to calculate the distance from the center to the foci of each hyperbola. For a hyperbola of the form [tex]\(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\)[/tex] or [tex]\(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\)[/tex], the focal distance [tex]\(c\)[/tex] is given by the formula [tex]\( c = \sqrt{a^2 + b^2} \)[/tex].
Let's go through each of the provided hyperbolas:
1. For equation A:
[tex]\[
\frac{(z-2)^2}{8^2}-\frac{(y-1)^2}{7^2}=1
\][/tex]
Here, [tex]\(a = 8\)[/tex] and [tex]\(b = 7\)[/tex], so:
[tex]\[
c_A = \sqrt{a^2 + b^2} = \sqrt{8^2 + 7^2} = \sqrt{64 + 49} = \sqrt{113} \approx 10.63
\][/tex]
2. For equation B:
[tex]\[
\frac{(2y+4)^2}{19^2}-\frac{(2z-6)^2}{11^2}=1
\][/tex]
Here, [tex]\(a = 19\)[/tex] and [tex]\(b = 11\)[/tex], so:
[tex]\[
c_B = \sqrt{a^2 + b^2} = \sqrt{19^2 + 11^2} = \sqrt{361 + 121} = \sqrt{482} \approx 21.95
\][/tex]
3. For equation C:
[tex]\[
\frac{(y-1)^2}{6^2}-\frac{(z-2)^2}{9^2}=1
\][/tex]
Here, [tex]\(a = 6\)[/tex] and [tex]\(b = 9\)[/tex], so:
[tex]\[
c_C = \sqrt{a^2 + b^2} = \sqrt{6^2 + 9^2} = \sqrt{36 + 81} = \sqrt{117} \approx 10.82
\][/tex]
4. For equation D:
[tex]\[
\frac{(y-5)^2}{5^2}-\frac{(2x+6)^2}{19^2}=1
\][/tex]
Here, [tex]\(a = 5\)[/tex] and [tex]\(b = 19\)[/tex], so:
[tex]\[
c_D = \sqrt{a^2 + b^2} = \sqrt{5^2 + 19^2} = \sqrt{25 + 361} = \sqrt{386} \approx 19.65
\][/tex]
Now, comparing all the computed [tex]\(c\)[/tex] values:
[tex]\[
c_A \approx 10.63, \quad c_B \approx 21.95, \quad c_C \approx 10.82, \quad c_D \approx 19.65
\][/tex]
The hyperbola with the greatest focal distance is equation B. Hence, the hyperbola that has its foci farthest from its center is represented by equation B:
[tex]\[
\boxed{\frac{(2y+4)^2}{19^2}-\frac{(2z-6)^2}{11^2}=1}
\][/tex]
