find an equation for the hyperbola vertices (3,0),(3,4) asymptotes:y=2/3,y=4-2/3x
center midway between vertices at (3,2)
slope of asymptote 2/3 = a/b
(y-2)^2/4 - (x-3)^2/9 = 1
thank you very much
To find the equation of a hyperbola, we need the coordinates of the vertices and the equations of the asymptotes.
Given:
Vertices: (3, 0), (3, 4)
Asymptotes: y = 2/3, y = 4 - (2/3)x
The general equation of a hyperbola with a vertical transverse axis is:
(x - h)^2 / a^2 - (y - k)^2 / b^2 = 1,
where (h, k) represents the coordinates of the center, 'a' is the distance from the center to the vertices along the transverse axis, and 'b' is the distance from the center to the co-vertices along the conjugate axis.
1. Determining the center:
The center of the hyperbola is the midpoint of the vertices. Since the x-coordinates of the vertices are the same (3), the x-coordinate of the center is also 3. Taking the average of the y-coordinates, we find that the y-coordinate of the center is (0 + 4) / 2 = 2.
So the center of the hyperbola is (3, 2).
2. Determining 'a':
The distance between the center and each vertex along the transverse axis is the same, so 'a' is the vertical distance from the center to either of the vertices, which in this case is 4 - 0 = 4.
3. Determining 'b':
The distance between the center and each co-vertex along the conjugate axis is given by the difference between the y-coordinate of the center and one of the asymptotes. We'll work with y = 2/3 to find 'b':
2 = 2/3 + b,
6/3 - 2/3 = b,
4/3 = b.
So 'b' is 4/3.
Now we can write the equation of the hyperbola:
(x - 3)^2 / 4^2 - (y - 2)^2 / (4/3)^2 = 1.
Simplifying,
(x - 3)^2 / 16 - (y - 2)^2 / 16/9 = 1.
Therefore, the equation of the hyperbola is:
(x - 3)^2 / 16 - (y - 2)^2 / (16/9) = 1.
To find the equation for the hyperbola with given vertices and asymptotes, you can follow these steps:
Step 1: Find the center of the hyperbola.
The center of a hyperbola is the midpoint between the vertices. In this case, the x-coordinate of the center is the same as the x-coordinate of both the vertices, which is 3. The y-coordinate of the center is the average of the y-coordinates of the vertices, which is (0 + 4)/2 = 2. Therefore, the center of the hyperbola is (3, 2).
Step 2: Find the distance between the center and one of the vertices.
Since the hyperbola is vertically aligned (the x-coordinates of the vertices are the same), the distance between the center and either of the vertices in the y-direction is the distance of the transverse axis. In this case, the distance is 4 - 2 = 2.
Step 3: Find the equation for the transverse axis.
The transverse axis of a hyperbola is the line passing through the center and the vertices. Since the hyperbola is vertically aligned, the equation of the transverse axis is x = 3.
Step 4: Find the slopes of the asymptotes.
The slopes of the asymptotes are given as y = 2/3 and y = 4 - (2/3)x. The slope-intercept form of a line is y = mx + b, where m is the slope of the line and b is the y-intercept. In this case, the slopes are 2/3 and -(2/3).
Step 5: Write the equation of the hyperbola.
The equation of a hyperbola with center (h, k), transverse axis length 2a, and slopes of asymptotes m1 and m2 is:
(x - h)^2 / a^2 - (y - k)^2 / b^2 = 1
Plugging in the given values, we have:
(x - 3)^2 / a^2 - (y - 2)^2 / b^2 = 1
Since the hyperbola is vertically aligned, the transverse axis length 2a is 2. Therefore, a = 1.
The slopes of the asymptotes are 2/3 and -(2/3), respectively. The distance from the center to each vertex in the y-direction is b.
Let's find b:
b = distance between the center and a vertex in the y-direction = 2.
Now we can plug in the values of a and b into the equation:
(x - 3)^2 / 1^2 - (y - 2)^2 / 2^2 = 1
Simplifying, we obtain the equation of the hyperbola:
(x - 3)^2 - (y - 2)^2 / 4 = 1