Asked by Ray Ray
Does Any One Have The Answers To...HyperBolas Quiz Part 1
please help its the end of the semester i'm just trying to get everything done
please help its the end of the semester i'm just trying to get everything done
Answers
Answered by
poopoohead
hey bot, write an equation of an ellipse in standard form with the center at the origin and with the given characteristics vertex at (-5,0) and covertex at (0,4)
Answered by
poopoohead
hey bot, A hyperbola has vertices (±5, 0) and one focus (6, 0). What is the standard-form equation of the hyperbola?
Answered by
Bot
As an AI language model, I do not have access to specific quizzes and their answers. However, I suggest reviewing your notes, textbook, and online resources to prepare for the quiz. You can also ask your teacher or a tutor for assistance. Good luck!
Answered by
Bot
Sure! The standard equation of an ellipse with center at the origin, vertex at (-5,0) and co-vertex at (0,4) is:
(x^2/25) + (y^2/16) = 1
In this equation, "a" is the distance from the center to the vertex, which is 5 in this case, and "b" is the distance from the center to the co-vertex, which is 4 in this case.
(x^2/25) + (y^2/16) = 1
In this equation, "a" is the distance from the center to the vertex, which is 5 in this case, and "b" is the distance from the center to the co-vertex, which is 4 in this case.
Answered by
Bot
Sure! To find the standard-form equation of the hyperbola with vertices (±5, 0) and focus (6, 0), we can first identify some important values:
Center: The center of the hyperbola is halfway between the vertices, so it is at (0, 0).
c: The distance from the center to a focus is c, which is 6 units in this case.
a: The distance from the center to a vertex is a, which is 5 units in this case.
Now we can use the standard form equation for a hyperbola, which is:
(x^2 / a^2) - (y^2 / b^2) = 1 (for a horizontal hyperbola)
or
(y^2 / a^2) - (x^2 / b^2) = 1 (for a vertical hyperbola)
Since the hyperbola has vertices on the x-axis and the focus is to the right of the center, we know that it is a horizontal hyperbola, and we can use the first equation.
Plugging in the values we identified, we get:
(x^2 / 5^2) - (y^2 / b^2) = 1
We still need to find b, which we can do using the formula:
b^2 = c^2 - a^2
Plugging in the values we identified, we get:
b^2 = 6^2 - 5^2
b^2 = 11
So the final standard-form equation of the hyperbola is:
(x^2 / 25) - (y^2 / 11) = 1.
Therefore, the standard-form equation of the hyperbola is (x^2 / 25) - (y^2 / 11) = 1.
Center: The center of the hyperbola is halfway between the vertices, so it is at (0, 0).
c: The distance from the center to a focus is c, which is 6 units in this case.
a: The distance from the center to a vertex is a, which is 5 units in this case.
Now we can use the standard form equation for a hyperbola, which is:
(x^2 / a^2) - (y^2 / b^2) = 1 (for a horizontal hyperbola)
or
(y^2 / a^2) - (x^2 / b^2) = 1 (for a vertical hyperbola)
Since the hyperbola has vertices on the x-axis and the focus is to the right of the center, we know that it is a horizontal hyperbola, and we can use the first equation.
Plugging in the values we identified, we get:
(x^2 / 5^2) - (y^2 / b^2) = 1
We still need to find b, which we can do using the formula:
b^2 = c^2 - a^2
Plugging in the values we identified, we get:
b^2 = 6^2 - 5^2
b^2 = 11
So the final standard-form equation of the hyperbola is:
(x^2 / 25) - (y^2 / 11) = 1.
Therefore, the standard-form equation of the hyperbola is (x^2 / 25) - (y^2 / 11) = 1.
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.