Equation for the hyperbola which has a vertices of (0;+_9) and asymtotes = y= +-18/5
Why did the hyperbola win the race? Because it had an unfair advantage with its speedy asymptotes, of course!
Now, let's find the equation for this hyperbola. The general equation for a hyperbola is:
((x - h)² / a²) - ((y - k)² / b²) = 1
where (h, k) are the coordinates of the center, and "a" and "b" are the distances from the center to the vertices along the x-axis and y-axis, respectively.
Given that the vertices are (0, ±9), we have a = 9.
Since the asymptotes are y = ±18/5, we know that the distance from the center to the asymptotes along the y-axis is b = 18/5.
Plugging in these values, the equation becomes:
(x² / 9²) - (y² / (18/5)²) = 1
Simplifying, we get:
(x² / 81) - (y² / 324/25) = 1
Multiplying through by the denominators and rearranging, we find:
25x² - 4y² = 81
So, the equation for the hyperbola with vertices (0, ±9) and asymptotes y = ±18/5 is 25x² - 4y² = 81.
To find the equation of a hyperbola with vertices and asymptotes, we can use the standard form equation for a hyperbola centered at the origin:
(x^2 / a^2) - (y^2 / b^2) = 1
Where a is the distance from the center to the vertices, and b is the distance from the center to the asymptotes.
Given that the vertices are (0, ±9) and the asymptotes are y = ±(18/5), we can determine the values of a and b:
The distance from the center to the vertices is a=9, so a^2=81.
The distance from the center to the asymptotes is b=18/5, so b^2=(18/5)^2 = 324/25.
Using these values, we can write the equation of the hyperbola as:
(x^2 / 81) - (y^2 / (324/25)) = 1
Multiplying both sides of the equation by 81 and 25 to eliminate fractions, we get:
25x^2 - 81y^2 = 81 * (324/25)
Simplifying:
25x^2 - 81y^2 = 324
Therefore, the equation for the hyperbola with vertices (0, ±9) and asymptotes y = ±(18/5) is:
25x^2 - 81y^2 = 324
To find the equation of a hyperbola given its vertices and asymptotes, you can follow these steps:
1. Determine the center of the hyperbola: The center is simply the midpoint between the two vertices. In this case, since the vertices are given as (0, +9) and (0, -9), the center will be at (0, 0).
2. Find the distance from the center to the vertices: The distance between the center and either vertex is called the semi-major axis. In this case, the semi-major axis is 9.
3. Determine the equation of the asymptotes: The equation of the asymptotes for a hyperbola with a vertical transverse axis (as in this case) is given by y = ±(b/a) * x, where "a" is the distance from the center to the vertices, and "b" is the distance from the center to the co-vertices. The co-vertices are the points on the transverse axis where the hyperbola intersects the asymptotes. From the given asymptotes, we can deduce that b, the distance from the center to the co-vertices, is 18/5.
4. Write the equation in standard form: The standard form of the equation of a hyperbola with a vertical transverse axis is (y - k)^2 / a^2 - (x - h)^2 / b^2 = 1, where (h, k) is the center of the hyperbola. Plugging in the values, we get:
(y - 0)^2 / 9^2 - (x - 0)^2 / (18/5)^2 = 1
Simplifying the equation further, we have:
y^2 / 81 - x^2 / (324/25) = 1
To get rid of the fraction in the denominator, we can multiply the entire equation by (25/324):
(25/324) * y^2 - x^2 = 25/324
Therefore, the equation of the hyperbola is:
25y^2 - 324x^2 = 25