Choose the equation that best represents an ellipse for the given foci and co-vertices.
1. foci (+-2, 0)
co-vertices (0, +-4)
2. foci (+-3, 0)
co-vertices (0,+-6)
3. foci (0, +-3)
co-vertices (+-5, 0)
Choose the equation for the hyperbola centered at the origin with the given characteristics.
8. one focus (0, square root of 34), one vertex (0, 5)
9. vertical transverse axis, b = 6, c = square root of 45
10. vertices (+-2, 0), perimeter of central rectangle 24 units
x^2/20 -y^2/16
To determine the equation that represents an ellipse with the given foci and co-vertices, you can use the standard form for the equation of an ellipse:
(x - h)^2/a^2 + (y - k)^2/b^2 = 1
where (h, k) represents the center of the ellipse, a represents the semi-major axis, and b represents the semi-minor axis.
Let's analyze each option:
1. For foci (+-2, 0) and co-vertices (0, +-4):
The center of the ellipse should be at the origin since the foci have the same x-coordinate and the co-vertices have the same y-coordinate, indicating symmetry around the y-axis. Thus, (h, k) = (0, 0).
The semi-major axis is equal to half the distance between the foci, which is 2 units.
The semi-minor axis is equal to half the distance between the co-vertices, which is 4 units.
Substituting these values into the standard form, we get:
x^2/4 + y^2/16 = 1
2. For foci (+-3, 0) and co-vertices (0, +-6):
Following a similar process as before, we find that:
(h, k) = (0, 0)
Semi-major axis = 3 units
Semi-minor axis = 6 units
The equation in standard form is:
x^2/9 + y^2/36 = 1
3. For foci (0, +-3) and co-vertices (+-5, 0):
This time, we have symmetry around the x-axis, indicating that the center of the ellipse is also at the origin: (h, k) = (0, 0).
Semi-major axis = 5 units
Semi-minor axis = 3 units
Putting these values into the standard equation form, we obtain:
x^2/25 + y^2/9 = 1
For the second part of your question, we will determine the equation for the hyperbola with the given characteristics.
8. For one focus (0, square root of 34) and one vertex (0, 5):
Since the vertex and the focus are on the y-axis, the hyperbola has a vertical transverse axis. This means the equation will be in the form:
(y - k)^2/a^2 - (x - h)^2/b^2 = 1
The center of the hyperbola is (h, k), which is the midpoint between the focus and the vertex. So, (h, k) = (0, (5 + sqrt(34))/2).
The distance from the center to the vertex (a) is 5 units.
The distance from the center to the focus (c) is the square root of 34 units.
Substituting the values into the standard equation form, we have:
(y - (5 + sqrt(34))/2)^2/25 - (x - 0)^2/b^2 = 1
9. For a vertical transverse axis, b = 6, and c = square root of 45:
Since we know b, we can calculate the value of a using the relationship a^2 = b^2 + c^2.
In this case, a^2 = 6^2 + sqrt(45)^2 = 36 + 45 = 81.
The equation in standard form is:
(y - k)^2/a^2 - (x - h)^2/b^2 = 1, where (h, k) represents the center of the hyperbola.
10. For vertices (+-2, 0) and a perimeter of the central rectangle of 24 units:
The perimeter of the central rectangle is given by the formula P = 2a + 2b, where a and b represent the semi-major and semi-minor axes, respectively.
In this case, P = 24, so we have 2a + 2b = 24.
Since the hyperbola is centered at the origin, (h, k) = (0, 0).
As we don't have enough information to determine a and b directly, we can proceed as follows:
We know that a vertex is located at (2, 0). Since the center is at the origin, the other vertex will be (-2, 0).
This distance is related to the semi-major axis, so a = 2 units.
Using the equation 2a + 2b = 24, we can substitute the value of a to find b:
2(2) + 2b = 24
4 + 2b = 24
2b = 24 - 4
2b = 20
b = 20/2
b = 10
Substituting these values into the standard equation form, we obtain:
x^2/4 - y^2/100 = 1