given the eccentilisty and a point that lies on the hyperbola ,find the equation of the hyperbola center origin

The name of your university is not the school subject you are asking about. Please put your school subject in the subject line so the proper tutor might find it.

To find the equation of a hyperbola with the center at the origin, given the eccentricity and a point that lies on the hyperbola, follow these steps:

Step 1: Identify the given information:
- Eccentricity (e)
- Coordinates of a point on the hyperbola (x1, y1)

Step 2: Determine the coordinates of the focus (F) using the eccentricity:
- The distance from the origin (O) to the focus (F) is given by the formula: |OF| = ae, where "a" is the distance between the center and a vertex.
- Since the center is at the origin (0, 0), the distance |OF| = ae simplifies to |F| = ae.
- Therefore, the coordinates of the focus are (ae, 0).

Step 3: Determine the coordinates of the given point (x1, y1).

Step 4: Calculate the distance from the point to the focus:
- The distance from the given point (x1, y1) to the focus (ae, 0) is given by the formula: √[(x1 - ae)^2 + (y1 - 0)^2].

Step 5: Use the distance formula for a hyperbola to create the equation:
[(x - h)^2 / a^2] - [(y - k)^2 / b^2] = 1, where (h, k) is the center of the hyperbola.

- Since the center is at the origin, the equation becomes:
x^2 / a^2 - y^2 / b^2 = 1

Step 6: Substitute the known values into the equation:
- Substitute the coordinates of the given point (x1, y1) and the coordinates of the focus (ae, 0) into the distance formula from Step 4.

- The equation should now have only three unknowns: a, b, and e.

Step 7: Solve for a, b, and e by simplifying the equation:
- Simplify the equation by expanding and rearranging it in terms of a, b, and e.

- You can then solve for a, b, and e using algebraic methods or by applying other known facts about the hyperbola (e.g., the relationship between a, b, and e).

By following these steps, you can find the equation of a hyperbola with the center at the origin, given the eccentricity and a point that lies on the hyperbola.