Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Discover detailed solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
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]
[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]
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.