what is the equation of a hyperbola with one vertex at (6,5) and the equation of asymptotes are 5x-6y-30 and 5x+6y-30?
To find the equation of a hyperbola with one vertex at (6,5) and the equation of asymptotes are 5x-6y-30 and 5x+6y-30, you can follow these steps:
Step 1: Determine the center of the hyperbola.
The center of the hyperbola is the midpoint between the two vertices. Since you have one vertex at (6,5), you need to find the other vertex to determine the center.
Step 2: Find the other vertex.
Since hyperbolas are symmetrical, the other vertex will have the same y-coordinate but a different x-coordinate. In this case, since one vertex is at (6,5), the other vertex will be at (-6,5).
Step 3: Determine the center.
The center is the midpoint between the two vertices. To find the midpoint, you can use the formula: (x₁ + x₂)/2, (y₁ + y₂)/2.
In this case, the center will be: (6 + (-6))/2, (5 + 5)/2.
Simplifying, the center is (0, 5).
Step 4: Find the values of a and b.
The values of a and b are related to the distance from the center to the vertices.
In this case, the distance from the center to either vertex, a, is 6 units.
Since the y-coordinate of the vertex doesn't change, the distance from the center to the transverse axis (or to the foci) is b = 5 units.
Step 5: Determine the equation of the hyperbola.
The equation of a hyperbola in standard form with center (h, k), transverse axis length 2a, and conjugate axis length 2b is given by:
[(x - h)² / a²] - [(y - k)² / b²] = 1.
Plugging in the values from the previous steps, you can write the equation of the hyperbola as follows:
[(x - 0)² / 6²] - [(y - 5)² / 5²] = 1.
Simplifying, the equation of the hyperbola is:
[x² / 36] - [(y - 5)² / 25] = 1.
Therefore, the equation of the hyperbola with one vertex at (6,5) and the equations of the asymptotes 5x-6y-30 and 5x+6y-30 is:
[x² / 36] - [(y - 5)² / 25] = 1.