To determine which hyperbola has its foci farthest from its center, we need to find the distance [tex]\(c\)[/tex] from the center to each focus for each given hyperbola equation. The general form of a hyperbola is:
[tex]\[
\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \quad \text{or} \quad \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1
\][/tex]
For a hyperbola, [tex]\(c\)[/tex] is given by:
[tex]\[
c = \sqrt{a^2 + b^2}
\][/tex]
Let's calculate [tex]\(c\)[/tex] for each hyperbola equation provided.
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].
[tex]\[
c = \sqrt{8^2 + 7^2} = \sqrt{64 + 49} = \sqrt{113} \approx 10.63
\][/tex]
Equation B:
[tex]\[
\frac{(2 y+4)^2}{19^2}-\frac{(2 x-6)^2}{11^2}=1
\][/tex]
Here, [tex]\(a = 19\)[/tex] and [tex]\(b = 11\)[/tex].
[tex]\[
c = \sqrt{19^2 + 11^2} = \sqrt{361 + 121} = \sqrt{482} \approx 21.95
\][/tex]
Equation C:
[tex]\[
\frac{(y-1)^2}{6^2}-\frac{(x-2)^2}{9^2}=1
\][/tex]
Here, [tex]\(a = 6\)[/tex] and [tex]\(b = 9\)[/tex].
[tex]\[
c = \sqrt{6^2 + 9^2} = \sqrt{36 + 81} = \sqrt{117} \approx 10.82
\][/tex]
Equation D:
[tex]\[
\frac{(y-5)^2}{5^2}-\frac{(2 x+6)^2}{10^2}=1
\][/tex]
Here, [tex]\(a = 5\)[/tex] and [tex]\(b = 10\)[/tex].
[tex]\[
c = \sqrt{5^2 + 10^2} = \sqrt{25 + 100} = \sqrt{125} \approx 11.18
\][/tex]
Comparing the values of [tex]\(c\)[/tex]:
[tex]\[
c_A \approx 10.63, \quad c_B \approx 21.95, \quad c_C \approx 10.82, \quad c_D \approx 11.18
\][/tex]
The hyperbola with the largest [tex]\(c\)[/tex] value is Equation B, as:
[tex]\[
c_B \approx 21.95
\][/tex]
Therefore, the correct answer is:
B. [tex]\(\frac{(2 y+4)^2}{19^2}-\frac{(2 x-6)^2}{11^2}=1\)[/tex]
