How do you find the center, vertices, and foci of the ellipse defined by the equation 2x² + 9y² = 18?

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To determine the center, vertices, and foci of the ellipse given by the equation 2x² + 9y² = 18, we need to first rewrite it in the standard form of an ellipse equation.

1. **Standard Form Conversion**:

Start by dividing all terms in the equation by 18 to make the right side equal to 1:

   2x²/18 + 9y²/18 = 1
   x²/9 + y²/2 = 1

This simplifies to:

   x²/9 + y²/(2/9) = 1

2. **Identifying Components**:

Now we can identify the values for a² and b² from the standard form:

  • From x²/9, we have a² = 9 so a = 3.
  • From y²/(2/9), we have b² = 2/9 so b = √(2/9) = √2/3.

3. **Finding the Center**:

The center of an ellipse in standard form (x-h)²/a² + (y-k)²/b² = 1 is at the point (h, k). In this case, since there are no additional constants subtracted or added in the equation, the center is at:

   (0, 0)

4. **Determining the Vertices**:

The vertices of the ellipse are located a distance a from the center along the major axis. Here, since 9 > 2, the major axis is along the x-axis. Thus, the vertices can be found at:

  • Vertices: (±a, 0) = (±3, 0)

5. **Finding the Foci**:

The foci can be found using the distance c, where c is determined by the formula c² = a² – b².

   c² = 9 - (2/9)
   c² = 9 - 0.222...
   c² = 8.777... = 79/9
   c = √(79/9) = √79/3

Thus the foci are located at:

  • Foci: (±c, 0) = (±√79/3, 0)

6. **Summary**:

  • Center: (0, 0)
  • Vertices: (3, 0) and (-3, 0)
  • Foci: (√79/3, 0) and (-√79/3, 0)

By following these steps, we can easily find the center, vertices, and foci of the given ellipse.

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