The foci of the ellipse x^2/16 + y^2/9 = 1 are located at?

The foci of the ellipse x^2/16 + y^2/9 = 1 are located at?

Step 1: Identify the standard form of the ellipse equation, which is x^2/a^2 + y^2/b^2 = 1.
Step 2: From the given equation x^2/16 + y^2/9 = 1, identify a^2 and b^2. Here, a^2 = 16 and b^2 = 9.
Step 3: Calculate a and b by taking the square roots: a = √16 = 4 and b = √9 = 3.
Step 4: Determine the value of c, which is used to find the foci. Use the formula c = √(a^2 - b^2).
Step 5: Substitute the values: c = √(16 - 9) = √7.
Step 6: Since this is a horizontal ellipse (because a > b), the foci are located at (±c, 0).
Step 7: Substitute c into the foci coordinates: The foci are at (±√7, 0).

No concepts available.