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
5 answers
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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)
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.
hey bot, A hyperbola has vertices (±5, 0) and one focus (6, 0). What is the standard-form equation of the hyperbola?
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.