SOLUTION: find the equation of hyperbola having center at the origin with transverse axis on the x-axis, a=4, and latus rectum 32

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: find the equation of hyperbola having center at the origin with transverse axis on the x-axis, a=4, and latus rectum 32      Log On


   



Question 1190985: find the equation of hyperbola having center at the origin with transverse axis on the x-axis, a=4, and latus rectum 32
Answer by CPhill(1959) About Me  (Show Source):
You can put this solution on YOUR website!
Here's how to find the equation of the hyperbola:
**1. General Equation:**
Since the hyperbola has its center at the origin and the transverse axis is along the x-axis, its general equation is:
(x²/a²) - (y²/b²) = 1
**2. Given Information:**
* a = 4 (distance from the center to each vertex along the transverse axis)
* Latus Rectum = 32
**3. Latus Rectum Formula:**
The length of the latus rectum is given by the formula:
Latus Rectum = (2b²) / a
**4. Solve for b²:**
We can plug in the given values and solve for b²:
32 = (2b²) / 4
32 = b²/2
b² = 32 * 2
b² = 64
**5. Write the Equation:**
Now that we have a² and b², we can write the equation of the hyperbola:
(x²/4²) - (y²/64) = 1
(x²/16) - (y²/64) = 1
Therefore, the equation of the hyperbola is (x²/16) - (y²/64) = 1.