Select the correct answer.

Which equation represents the hyperbola that has its foci farthest from its center?

A. [tex]\frac{(z-2)^2}{8^2}-\frac{(y-1)^2}{7^2}=1[/tex]

B. [tex]\frac{(2y+4)^2}{19^2}-\frac{(2x-6)^2}{11^2}=1[/tex]

C. [tex]\frac{(y-1)^2}{6^2}-\frac{(x-2)^2}{9^2}=1[/tex]

D. [tex]\frac{(y-5)^2}{5^2}-\frac{(2x+6)^2}{10^2}=1[/tex]



Answer :

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]