equation for the hyperbola which has vertices (0;9) and asymptotes y= 18/5 x
from the asymptote
b/a = 18/5 , but b = 9
b/a = 9/2.5
x^2/(2.5)^2 - y^2/81 = -1
To find the equation of a hyperbola given its vertices and asymptotes, we can use the standard form equation for a hyperbola:
(x-h)²/a² - (y-k)²/b² = 1
Where (h,k) represent the center of the hyperbola, and 'a' and 'b' represent the distances from the center to the vertices along the x-axis and y-axis, respectively.
In this case, we are given the vertices (0, 9) and asymptotes y = (18/5)x.
Step 1: Determine the center of the hyperbola
The center of the hyperbola is the midpoint between the vertices. In this case, since one vertex is at (0, 9), the center will also have a y-coordinate of 9. To find the x-coordinate, we can take the average of the x-coordinates of the vertices, which is (0 + 0)/2 = 0. Therefore, the center of the hyperbola is (0, 9).
Step 2: Determine the distance from the center to the vertices along the x-axis (a)
The distance from the center to the vertices along the x-axis is the absolute value of the x-coordinate of one of the vertices. In this case, since both vertices have an x-coordinate of 0, the distance is 0. Therefore, a = 0.
Step 3: Determine the distance from the center to the asymptotes along the y-axis (b)
The distance from the center to the asymptotes along the y-axis can be found using the formula: b = |(slope of asymptotes)|/sqrt(a² + b²). In this case, the slope of the asymptotes is 18/5, and a = 0. Therefore, b = |(18/5)|/sqrt(0² + b²). Since a = 0, the equation simplifies to b = 18/5.
Step 4: Substitute all the values (center, a, and b) into the equation
Plugging the values into the standard form equation, we get:
(x-0)²/0² - (y-9)²/(18/5)² = 1
Simplifying further, we have:
x²/(0) - (y-9)²/(324/25) = 1
As we can see, the term (x-0)²/0² becomes undefined, and thus, the equation cannot be determined. This suggests that there is a mistake or inconsistency in the given information. Please double-check the given values to ensure their accuracy.